Hc Verma I for Class 11 Science Physics Chapter 11 - Gravitation

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Page No 223:

Answer:

A particle will be in equilibrium when the net force acting on it is equal to zero. Two particles under the action of their mutual gravitational force will be in equilibrium when they revolve around a common point under the influence of their mutual gravitational force of attraction. In this case, the gravitational pull is used up in providing the necessary centripetal force. Hence, the net force on the particles is zero and they are in equilibrium. This is also true for a three particle system.

Page No 223:

Answer:

Weight of a body is always because of its gravitational attraction with earth.As law of gravitational attraction is universal so it applies to any two bodies (earth and sun as well).So we can define the weight of earth w.r.t a body of mass comparable or heavier than earth because of its gravitational attraction between earth and that body.
But practically no body on earth has mass comparable to earth so weight of earth will be a meaningless concept w.r.t earth frame.

Page No 223:

Answer:

We know that acceleration due to a force on a body of mass in given by $a = \frac{F}{m}$ .
If F is the gravitational force acting on a body of mass m, then a is the acceleration of a free falling body.

This force is given as $F = G \frac{M m}{R^{2}}$ .
Here, M is the mass of the Earth; G is the universal gravitational constant and R is the radius of the Earth.
∴ Acceleration due to gravity, $a = \frac{F}{m} = \frac{G M}{R^{2}}$
From the above relation, we can see that acceleration produced in a body does not depend on the mass of the body. So, acceleration due to gravity is the same for all bodies.

Page No 223:

Answer:

No. All practicals which have mass exert gravitational force on each other. Even massless particles experience the same gravitational force like other particles, because they do have relativistic mass.

Page No 223:

Answer:

We know that the Earth-Moon system revolves around the Sun. The gravitational force of the Sun on the system provides the centripetal force its revolution. Therefore, the net force on the system is zero and the Moon does not experience any force from the Sun. This is the reason why the Moon revolves around the Earth and not around the Sun.

Page No 223:

Answer:

No. Due to the revolution of the Earth around the Sun, the gravitational force of the Sun on the Earth system is almost zero. Hence, the body will not experience any force due to the Sun. Therefore, weight of the object will remain the same.

Page No 223:

Answer:

The mutual gravitational force between the apple and the Earth is responsible for the acceleration produced in the apple falling from the tree. Although the Earth will experience the same force, it does not get attracted towards the apple because of its large mass. The insect feels that the Earth is falling towards the apple with an acceleration g because of the the relative motion.
let
v_ae=velocity of apple w.r.t earth
$V_{e a} = v e l o c i t y o f e a r t h w . r . t a p p l e$
$v_{a e} = v_{a} - v_{e} = - (v_{e} - v_{a}) = - v_{e a}$
As the insect is in the frame of apple so he sees the earth moving with a relative velocity $v_{e a}$ .
Any other observer on earth will see the apple moving towards earth with velocity $v_{a e}$ .Both are opposite in direction.

Page No 223:

Answer:

The gravitational potential due to the system is given as $V = \frac{k}{r^{2}}$ .
Gravitational field due to the system:
$E = - \frac{d V}{d r} \Rightarrow E = - \frac{d}{d r} (\frac{k}{r^{2}}) = - (- \frac{2 k}{r^{3}}) \Rightarrow E = \frac{2 k}{r^{3}}$
We can see that for this system, $E \propto \frac{1}{r^{3}}$
This type of system is not possible because $F_{g}$ is always proportional to inverse of square of distance(experimental fact).

If there were negative masses, then this type of system is possible.

This system is a dipole of two masses, i.e., two masses, one positive and the other negative, separated by a small distance.

In this case, the gradational field due to the dipole is proportional to $\frac{1}{r^{3}}$ .

Page No 223:

Answer:

The gravitational potential energy of a two-particle system is given by $U = - \frac{G m_{1} m_{2}}{r}$ .
This relation does not tell that the gravitational potential energy is zero at infinity. For our convenience, we choose the potential energies of the two particles to be zero when the separation between them is infinity.

No, if we suppose that the potential energy for $r = \infty$ is 20 J, then we need to modify the formula.

Now, potential energy of the two-particle system separated by a distance r is given by

$U (r^{,}) = U (r) - U (\infty) Given : U (\infty) = 20 J ∴ U (r) = - G \frac{m_{1} m_{2}}{r^{2}} - 20$

This formula should be used to calculate the gravitational potential energy at separation r.

Page No 223:

Answer:

The weight of an object is more at the poles than that at the equator. In purchasing or selling goods, we measure the mass of the goods. The balance used to measure the mass is calibrated according to the place to give its correct reading. So, it is not beneficial to purchase goods at the equator and sell them at the poles. A beam balance measures the mass of an object, so it can be used here. For using a spring balance, we need to calibrate it according to the place to give the correct readings.

Page No 223:

Answer:

The weight of a body at the poles is greater than that at the equator. Here, we are talking about the actual weight of the body at that particular place.
Yes. If the rotation of the Earth is taken into account, then we are discussing the apparent weight of the body.

Page No 224:

Answer:

If we consider the Earth to be a perfect sphere, then the acceleration due to gravity at its surface is given by $g = G \frac{M}{R^{2}}$ .
Here, M is the mass of Earth; R is the radius of the Earth and G is universal gravitational constant.

If the radius of the earth is decreased by 1%, then the new radius becomes

$R' = R - \frac{R}{100} = \frac{99}{100} R \Rightarrow R' = 0.99 R$

New acceleration due to gravity will be given by
$g' = G \frac{M}{R'^{2}} = G \frac{M}{(0.99 R)^{2}} \Rightarrow g' = 1.02 \times (G \frac{M}{R^{2}}) = 1.02 g$
Hence, the value of the acceleration due to gravity increases when the radius is decreased.

Percentage increase in the acceleration due to gravity is given by
$\frac{g' - g}{g} \times 100 = \frac{0.02 g}{g} \times 100 = 2 %$

Page No 224:

Answer:

No, it will not land on the Earth. The nut will start revolving in the orbit of the satellite with the same orbital speed as that of the satellite due to inertia of motion. An astronaut can make it land on the Earth by projecting it with some velocity toward the Earth.

Page No 224:

Answer:

According to Kepler first law of planetary motion all planets move in elliptical orbits with sun at one of its foci. It applies to any planet and its satellite as well.This implies that plane of the satellite has to pass through the centre of planet (earth).

Page No 224:

Answer:

$T=(\frac{g R^{2} T^{2}}{4 π^{2}})^{\frac{1}{3}} - R T = \frac{4 π^{2} (h + R)^{3}}{g R^{2}} = \frac{4 \times 3 . 14^{2} \times (36000 + 6400)^{3} \times 10^{9}}{9.8 \times (6400 \times 10^{3})^{2}} = 24.097 Hr Which implies that it is a geostationary sattelite with time period=24 Hrs.$

Page No 224:

Answer:

No. All geostationary orbits are concentric with the equator of the Earth.

Page No 224:

Answer:

A person living in a house at the equator will not feel weightlessness because he is not in a free fall motion. Satellites are in free fall motion under the gravitational pull of the earth, but, due to the curved surface of the Earth, they move in a circular path. The gravitational force on the satellite due to the Sun provides the centripetal force for its revolution. Therefore, net force on the satellite is zero and, thus, a person feels weightless in a satellite orbiting the earth.

Page No 224:

Answer:

No, both satellites will have different time periods as seen from the Earth. The satellite moving opposite (east to west) to the rotational direction of the Earth will have less time period, because its relative speed with respect to the Earth is more.

Page No 224:

Answer:

Yes, a spacecraft consumes more fuel in going from the Earth to the Moon than it takes for the return trip. In going from the Earth to the Moon, the spacecraft has to overcome the gravitational pull of the earth. So, more fuel is consumed in going from the Earth to Moon. However, in the return trip, this gravitation pull helps the spacecraft to come back to the Earth.

Page No 224:

Answer:

(b) 0⋅0027 m s⁻²

We know that the distance of the Moon from the Earth is about 60 times the radius of the earth. So, acceleration due to gravity at that distance is 0.0027 m/s². When the Moon is stopped for an instant and then released, it will fall towards the Earth with an initial acceleration of 0.0027 m/s².

Page No 224:

Answer:

(c) 6⋅4 m s⁻²

According to the previous question, we have:
Radius of the moon, $R_{m} = \frac{R_{e}}{4} = \frac{6400000}{4} = 1600000 m$
So, when the Moon is just about to hit the surface of the Earth, its centre of mass is at a distance of (R_e + R_m) from the centre of the Earth.

Acceleration of the Moon just before hitting the surface of the earth is given by

$g' = \frac{G M}{(R_{e} + R_{m})^{2}} = \frac{G M}{{R_{e}}^{2} (1 + \frac{R_{m}}{R_{e}})^{2}} \Rightarrow g' = \frac{g}{(1 + \frac{R_{m}}{R_{e}})^{2}} = \frac{10}{(1 + \frac{1}{4})^{2}} = \frac{10 \times 16}{25} \Rightarrow g' = 6.4 m / s^{2}$

Page No 224:

Answer:

(c) C

At one point between the Earth and Mars, the gravitational field intensity is zero. So, at that point, the weight of the passenger is zero. The curve C indicates that the weight of the passenger is zero at a point between the Earth and Mars.

Page No 224:

Answer:

(d) 2^1/3 W at the planet

The weight of the object on the Earth is $W = m \frac{G M_{e}}{{R_{e}}^{2}}$ .
Here, m is the actual mass of the object; M_eis the mass of the earth and R_e is the radius of the earth.
Let R_p be the radius of the planet.
Mass of the planet, $M_{p} = 2 M_{e}$
If $ρ$ is the average density of the planet then

$\frac{4}{3} π {R_{p}}^{3} \times ρ = 2 \times (\frac{4}{3} π {R_{e}}^{3} \times ρ) \Rightarrow R_{p} = {(2)}^{\frac{1}{3}} R_{e}$

Now, weight of the body on the planet is given by
$W_{p} = m (\frac{G M_{p}}{{R_{p}}^{2}}) = m (\frac{2 G M_{e}}{2^{\frac{2}{3}} {R_{e}}^{2}}) \Rightarrow W_{p} = 2^{\frac{1}{3}} \times m (\frac{G M_{e}}{{R_{e}}^{2}}) \Rightarrow W_{p} = 2^{\frac{1}{3}} \times W$

Page No 224:

Answer:

(a) $\frac{1}{2} m g R$

Work done = $-$ (final potential energy $-$ initial potential energy)
$\Rightarrow W = - (\frac{G M m}{2 R} - \frac{G M m}{R}) \Rightarrow W = \frac{1}{2} \frac{G M m}{R} = \frac{1}{2} m R \times (\frac{G M}{R^{2}}) \Rightarrow W = \frac{1}{2} m R g [∵ g = \frac{G M}{R^{2}}]$

Page No 224:

Answer:

The work done by the person is equal to the kinetic energy and the potential energy of the mass of 1 kg at point A.

Let V_A be the potential at point A.

$Now, W = \frac{1}{2} m v^{2} + {(P . E .)}_{A} \Rightarrow W = \frac{1}{2} m v^{2} + V_{A} \times m \Rightarrow - 3 = \frac{1}{2} \times 1 \times (2)^{2} + V_{A} \times 1 \Rightarrow V_{A} = - 5 J / Kg$

Page No 224:

Answer:

(c) B is correct but A is wrong.

The plot of E against r is discontinuous as gravitational field inside the spherical shell is zero (r < R). The plot of V against r is a continuous curve for a uniform spherical shell.

Page No 224:

Answer:

(d) Both A and B are wrong.

Both the plots (i.e., V against r and E against r) are continuous curves for a uniform solid sphere.

Page No 224:

Answer:

(b) A is correct but B is wrong.

We know that the value of acceleration due to gravity decreases when we go up from the surface of the Earth. If we take the into account the effect of bulging of the Earth due to its rotation, we can say that acceleration due to gravity is maximum at the poles and minimum at the equator

So, there are points above both the poles where the value of g is equal to its value at the equator.

Page No 225:

Answer:

The gravitational force of attraction between the balls is given as by
$F = \frac{G m_{1} m_{2}}{r^{2}} Given : m_{1} = m_{2} = 10 kg and r = 10 cm = 0.10 m$
$∴ F = \frac{6.67 \times 10^{- 11} \times 10 \times 10}{{(0.1)}^{2}} \Rightarrow F = 6.67 \times 10^{- 7} N$

Page No 225:

Answer:

$Force due to the particle at A, {\vec{F}}_{OA} = \frac{G \times m \times m}{{OA}^{2}} Let OA = r ∴ {\vec{F}}_{OA} = \frac{G \times m \times m}{r^{2}} Here, r = \sqrt{{(\frac{a}{2})}^{2} + {(\frac{a}{2})}^{2}} = \frac{a}{\sqrt{2}} Force due to the particle at B, {\vec{F}}_{OB} = \frac{G \times m \times 2 m}{r^{2}} Force due to the particle at C, {\vec{F}}_{OC} = \frac{G \times m \times 3 m}{r^{2}} Force due to the particle at D, {\vec{F}}_{OD} = \frac{G \times m \times 4 m}{r^{2}} Now, resultant force = {\vec{F}}_{OA} + {\vec{F}}_{OB} + {\vec{F}}_{OC} + {\vec{F}}_{OD} = \frac{2 G m m}{a^{2}} [- \frac{\vec{i}}{\sqrt{2}} + \frac{\vec{j}}{\sqrt{2}}] + \frac{4 G m m}{a^{2}} [\frac{\vec{i}}{\sqrt{2}} + \frac{\vec{j}}{\sqrt{2}}] = \frac{6 G m m}{a^{2}} [\frac{\vec{i}}{\sqrt{2}} - \frac{\vec{j}}{\sqrt{2}}] + \frac{8 G m m}{a^{2}} [\frac{- \vec{i}}{\sqrt{2}} - \frac{- \vec{j}}{\sqrt{2}}] ∴ F = \frac{4 \sqrt{4} G m^{2}}{a^{2}} \overset{⏜}{j}$

Page No 225:

Answer:

(a) the mass of the satellite

The time period of an earth-satellite in circular orbit is independent of the mass of the satellite, but depends on the radius of the orbit.

Page No 225:

Answer:

(b) U > K

For a system to be bound , total energy of the system should be negative.As we know that kinetic energy can never be negative.
E<0 & K>0 & E=K+U
it gives that U>K.

Page No 225:

Answer:

(b) t₁ = t₂

Kepler's second law states that the line joining a planet to the Sun sweeps out equal areas in equal intervals of time, i.e., areal velocity of a planet about the Sun is constant.

The given areas swept by the planet are equal, so t₁ = t₂

Page No 225:

Answer:

(c) the normal force is zero

The gravitational pull on the satellite is used up in providing the necessary centripetal force required for its revolution around the earth. This means that there is no net force on the person sitting in a chair in the satellite. So, the normal reaction of the chair on the person is zero and he will feel weightless.

Page No 225:

Answer:

(a) W₁ = W₂

The gravitational pull on the satellite in both cases is used up in providing the necessary centripetal force required for its revolution around the earth. This means that there is no net force acting on the body which has been suspended from a spring balance in the satellite. So, the readings of the spring balance in both the cases are the same and is equal to zero.

Page No 225:

Answer:

(c) $m g R$

The kinetic energy needed to project a body of mass m from the Earth's surface to infinity is equal to the negative of the change in potential energy of the body.

i.e., kinetic energy = $-$ (final potential energy $-$ initial potential energy)
$\Rightarrow K = - (\frac{G M m}{R'} - \frac{G M m}{R}) \Rightarrow K = - (\frac{G M m}{\infty} - \frac{G M m}{R}) = \frac{G M m}{R} \Rightarrow K = m R \times (\frac{G M}{R^{2}}) \Rightarrow K = m R g [∴ g = \frac{G M}{R^{2}}]$

Page No 225:

Answer:

(c) $\sqrt{\frac{G M}{R}}$

Potential energy of the particle at a distance R from the surface of the Earth is ${(P . E .)}_{i} = \frac{G M m}{(R + R)} = \frac{1}{2} \frac{G M m}{R}$ .
Here, M is the mass of the earth; R is the radius of the earth and m is the mass of the body.

Let the particle be projected with speed v so that it just escapes the gravitational pull of the earth.
So, kinetic energy of the body = $-$ [change in the potential energy of the body]

Now, kinetic energy of the body = $-$ [final potential energy $-$ initial potential energy]
$\Rightarrow \frac{1}{2} m v^{2} = - [\frac{G M m}{\infty} - \frac{G M m}{2 R}] \Rightarrow v = \sqrt{\frac{G M m}{R}}$

Page No 225:

Answer:

(d) will depend on the direction of projection.For example a body projected vertically requires less escape velocity than a body projected at an angle with the vertical.

Page No 225:

Answer:

(a) $V = 0 and E = 0$
(b) $V = 0 and E \neq 0$
(c) $V \neq 0 and E = 0$
(d) $V \neq 0 and E \neq 0$ .

All the given conditions for gravitational potential and gravitational field at a point are possible.

Page No 225:

Answer:

(b) the gravitational field is zero
(c) the gravitational potential is same everywhere
(d) the gravitational field is same everywhere

Inside a uniform spherical shell, the gravitational field is the same everywhere and is equal to zero. The gravitational potential has a constant value inside a uniform spherical shell.

Page No 225:

Answer:

(b) decreases
AS it is maintaining its shape so its mass will remain constant which implies that if it shrinks then its volume decreases and to keep the mass constant ,its density increases.Also the gravitational potential at the centre of a uniform spherical shell is inversely proportional to the radius of the shell with a negative sign. When a uniform spherical shell gradually shrinks, the gravitational potential at the centre decreases (because of the negative sign in the formula of potential).

Page No 225:

Answer:

(b) is zero in some parts of the orbit
(c) is zero in one complete revolution

When a planet is moving in an elliptical orbit, at some point, the line joining the centre of the Sun and the planet is perpendicular to the velocity of the planet. For that instant, work done by the gravitational force on the planet becomes zero. As there is no net increase in the speed of the planet after one complete revolution about the Sun, the work done by the gravitational force on the planet in one complete revolution is zero.
Note:For elliptical orbits angle between force ans velocity is always 90 so there the work done is zero in any small part of the orbit.

Page No 225:

Answer:

(a) Speeds of A and B are equal.

The orbital speed of a satellite is independent of the mass of the satellite, but it depends on the radius of the orbit. Potential energy, kinetic energy and total energy depend on the mass of the the satellite.

Page No 225:

Answer:

(d) Angular momentum.

In planetary motion, the net external torque on the planet is zero. Therefore, angular momentum will remain constant.

Page No 226:

Answer:

(a)
Consider that mass 'm' is placed at the midpoint O of side AB of equilateral triangle ABC.

AO = BO = $\frac{a}{2}$

Then ${\vec{F}}_{OA} = \frac{4 G m^{2}}{a^{2}}$ along OA
Also, ${\vec{F}}_{OB} = \frac{4 G m^{2}}{a^{2}}$ along OB
OC = $\frac{\sqrt{3} a}{2}$
${\vec{F}}_{OC} = \frac{4 G m^{2}}{\{(3) a^{2}\}} = \frac{4 G m^{2}}{3 a^{2}}$ along OC

The net force on the particle at O is $\vec{F} = {\vec{F}}_{OA} + {\vec{F}}_{OB} + {\vec{F}}_{OC}$ .

Since equal and opposite forces cancel each other, we have:
$\vec{F} = {\vec{F}}_{OC} = \frac{4 G m^{2}}{\{(3) a^{2}\}} = \frac{4 G m^{2}}{3 a^{2}}$ along OC.

(b) If the particle placed at O (centroid)

All the forces are equal in magnitude but their directions are different as shown in the figure.

Equal and opposite forces along OM and ON cancel each other.

i.e., $F \cos 30 ° = F \cos 30 °$

∴ Resultant force $= F - 2 F \sin 30 = 0$

Page No 226:

Answer:

Three spheres are placed with their centres at A, B and C as shown in the figure.

Gravitational force on sphere C due to sphere B is given by
${\vec{F}}_{CB} = \frac{G m^{2}}{4 a^{2}} \cos 60 ° \hat{i} + \frac{G m^{2}}{4 a^{2}} \cdot \sin 60 ° \hat{j}$

Gravitational force on sphere C due to sphere A is given by

${\vec{F}}_{CA} = - \frac{G m^{2}}{4 a^{2}} \cos 60 ° \hat{i} + \frac{G m^{2}}{4 a^{2}} \cdot \sin 60 ° \hat{j}$

$∴ {\vec{F}}_{CB} = {\vec{F}}_{CB} + {\vec{F}}_{CA} = + \frac{2 G m^{2}}{4 a^{2}} \sin 60 ° \hat{j} = + \frac{2 G m^{2}}{4 a^{2}} \times \frac{\sqrt{3}}{2}$
i.e., magnitude $= \frac{\sqrt{3} G m^{2}}{4 a^{2}}$ along CO

Page No 226:

Answer:

Assume that three particles are at points A, B and C on the circumference of a circle.
BC = CD = $\sqrt{2} a$

The force on the particle at C due to gravitational attraction of the particle at B is ${\vec{F}}_{CB} = \frac{G M^{2}}{2 R^{2}} \hat{j}$ .
The force on the particle at C due to gravitational attraction of the particle at D is ${\vec{F}}_{CD} = - \frac{G M^{2}}{2 R^{2}} \hat{i}$ .
Now, force on the particle at C due to gravitational attraction of the particle at A is given by

${\vec{F}}_{CA} = - \frac{G M^{2}}{4 R^{2}} \cos 45 \hat{i} + \frac{G M^{2}}{4 R^{2}} \sin 45 \hat{j} ∴ {\vec{F}}_{C} = {\vec{F}}_{CA} + {\vec{F}}_{CB} + {\vec{F}}_{CD} = \frac{- G M^{2}}{4 R^{2}} (2 + \frac{1}{\sqrt{2}}) \hat{i} + \frac{G M^{2}}{4 R^{2}} (2 + \frac{1}{\sqrt{2}}) \hat{j}$

So, the resultant gravitational force on C is $F_{C} = \frac{G m^{2}}{4 R^{2}} \sqrt{2 \sqrt{2} + 1}$ .

Let v be the velocity with which the particle is moving.
Centripetal force on the particle is given by
$F = \frac{m v^{2}}{R} \Rightarrow v = \sqrt{\frac{G M}{R} (\frac{2 \sqrt{2} + 1}{4})}$

Page No 226:

Answer:

The acceleration due to gravity at a point at height h from the surface of the moon is given by
$g = \frac{G M}{r^{2}}$ ,
where M is the mass of the moon; r is the distance of point from the centre of the moon and G is universal gravitational constant.

$∴ g = \frac{G M}{{(R + h)}^{2}} \Rightarrow g = \frac{6.67 \times 10^{- 11} \times 7.4 \times 10^{22}}{{(1740 + 1000)}^{2} \times 10^{6}} \Rightarrow g = \frac{6.67 \times 7.4 \times 10^{11}}{(1740 + 1000^{2} \times 10^{6})} \Rightarrow g = \frac{6.67 \times 7.4 \times 10^{11}}{2740 \times 2740 \times 10^{6}} \Rightarrow g = 0.65 m / s^{2}$

Page No 226:

Answer:

Consider a system of two bodies. The initial linear momentum of the system is zero as the bodies were initially at rest when they were released.
Since the gravitational force is an internal force and the net external force on the system is zero, so by the law of conservation of linear momentum, the final momentum of the system will also be zero.

So, 10 $\times$ v₁ = 20 $\times$ v₂
⇒ v₁ = 2v₂ ...(i)
Applying the law of conservation of energy, we have:

Initial total energy = final total energy ...(ii)

Initial total energy $= \frac{- 6.67 \times 10^{11} \times 10 \times 20}{1}$ + 0
= −13.34 × 10⁻⁹ J ...(iii)

When the separation is 0.5 m, we have:

Final total energy $= \frac{- 13.34 \times 10^{- 9}}{1 / 2} + (\frac{1}{2}) \times 10 v_{1}^{2} + (\frac{1}{2}) \times 20 v_{2}^{2} . . . (i v)$
From (iii) and (iv), we have:
−13.34 × 10⁻⁹ = 26.68 × 10⁻⁹ + $5 v_{1}^{2} + 10 v_{2}^{2}$
⇒−13.34 × 10⁻⁹ = 26.68 + 10⁻⁹ + $30 v_{2}^{2}$
$\Rightarrow v_{2}^{2} = \frac{- 13.34 \times 10^{- 9}}{30}$ = 4.44 × 10⁻¹⁰
⇒ v₂ = 2.1 × 10⁻⁵ m/s
∴ v₁ = 4.2 × 10⁻⁵ m/s

Page No 226:

Answer:

Consider a small mass element of length dl subtending dθ angle at the centre.

In the semicircle, we can consider a small element dθ.
Then length of the element, dl = R dθ
Mass of the element, dm $= (\frac{M}{L}) R d θ$
Force on the mass element is given by
$d F = \frac{G m}{R^{2}} d m = \frac{G M R m}{L R^{2}} d θ$
The symmetric components along AB cancel each other.
Now, net gravitational force on the particle at O is given by

$F = \int 2 d F \sin θ = \int \frac{2 G M m}{L R} \sin θ d θ ∴ F = \int_{0}^{π / 2} \frac{- 2 G M m}{L R} \sin θ d θ = \frac{2 G M m}{L R} {[- \cos θ]}_{0}^{π / 2} = - 2 \frac{G M m}{L R} (- 1) = \frac{2 G M m}{L R} = \frac{2 G M m}{L L / π} = \frac{2 πG M m}{L^{2}}$

Page No 226:

Answer:

Consider a small mass element of length dx at a distance x from the centre of the rod.
Mass of the mass element, dm = (M/L) × dx

Gravitational field due to this element at point P is given by

$d E = \frac{G (d m) \times 1}{(d^{2} + x^{2})}$

The components of the gravitational field due to the symmetrical mass element along the length of the rod cancel each other.

Now, resultant gravitational field = 2dE sin θ
$= 2 \times \frac{G (d m)}{(d^{2} + x^{2})} \times \frac{d}{\sqrt{(d^{2} + x^{2})}} = \frac{2 \times G M \times d d x}{L (d^{2} + x^{2}) \{(\sqrt{d^{2} + x^{2}})\}}$
Total gravitational field due to the rod at point P is given by
$E = \int_{0}^{L / 2} \frac{2 G m d d x}{L {(d^{2} + x^{2})}^{3 / 2}}$ .
On integrating the above equation, we get:
$E = \frac{2 G m}{d \sqrt{L^{2} + 4 d^{2}}}$

Page No 226:

Answer:

Consider that mass m is at a distance $\frac{R_{1} + R_{2}}{2}$ as shown in the figure.

The gravitational force of m due to the shell of M₂ is zero, because the mass is inside the shell.

∴ Gravitational force due to the shell of mass M₂ = $\frac{G M_{1} m}{{(\frac{R_{1} + R_{2}}{2})}^{2}} = \frac{4 G M_{1} m}{{(R_{1} + R_{2})}^{2}}$

Page No 226:

Answer:

Mass of the Earth, $M = (\frac{4}{3}) π R^{3} ρ$ ...(i)
Consider an imaginary sphere of radius x with centre O as shown in the figure below:

$Mass of the imaginary sphere, M' = (\frac{4}{3}) π x^{3} ρ . . . (i i) From (i) and (i i), we have : \frac{M'}{M} = \frac{x^{3}}{R^{3}}$

∴ Gravitational force on the particle of mass m is given by
F $= \frac{G M m}{x^{2}}$
$\Rightarrow F = \frac{G M x^{3} m}{R^{3} x^{2}} = \frac{G M m}{R^{3}} x$

Page No 226:

Answer:

Let d be distance of the particle from the centre of the Earth.

$Now, d^{2} = x^{2} + (\frac{R^{2}}{4}) = \frac{4 x^{2} + R^{2}}{4} \Rightarrow d = (\frac{1}{2}) \sqrt{4 x^{2} + R^{2}}$
Let M be the mass of the Earth and M' be the mass of the sphere of radius d.
Then we have:
$M = (\frac{4}{3}) π R^{3} ρ M^{1} = (\frac{4}{3}) π d^{3} ρ ∴ \frac{M^{1}}{M} = \frac{d^{3}}{R^{3}}$

Gravitational force on the particle of mass m is given by
$F = \frac{G M^{1} m}{d^{2}} \Rightarrow F = \frac{G d^{3} M m}{R^{3} d^{2}} \Rightarrow F = \frac{G M m}{R^{3}} d$
∴ Normal force exerted by the wall, F_N = F cos θ $= \frac{GM m d}{R^{3}} \times \frac{R}{2 d} = \frac{GM m d}{2 R^{2}}$

Page No 226:

Answer:

(a) Consider that the particle is placed at a distance x from O.

Here, r < x < 2r
Let us consider a thin solid sphere of radius (x $-$ r).

Mass of the sphere, $d m = \frac{m}{(\frac{4}{3}) π r^{3}} \times \frac{4}{3} π (x - r)^{3} = \frac{m (x - r)^{3}}{r^{3}}$

Then the gravitational force on the particle due to the solid sphere is given by
$F = \frac{G m' d m}{(x - r)^{2}} = \frac{G \frac{m (x - r)^{3}}{r^{3}} m'}{(x - r)^{2}} = \frac{G m m' (x - r)}{r^{3}}$
Force on the particle due to the shell will be zero because gravitational field intensity inside a shell is zero.

(b) If 2r < x < 2R,
Force on the body due to the shell will again be zero as particle is still inside the shell.
then F is only due to the solid sphere.
$∴ F = \frac{G m m'}{{(x - r)}^{2}}$

(c) If x > 2R, then the gravitational force is due to both the sphere and the shell.
Now, we have:
Gravitational force due to shell, $F = \frac{GM m'}{{(x - R)}^{2}}$
Gravitational force due to the sphere $= \frac{G m m'}{{(x - r)}^{2}}$
As both the forces are acting along the same line joining the particle with the centre of the sphere and shell so both the forces can be added directly without worrying about their vector nature.
∴ Resultant force $= \frac{G m m'}{{(x - r)}^{2}} + \frac{G M m'}{{(x - R)}^{2}}$

Page No 226:

Answer:

At point P₁, the gravitational field due to the sphere and the shell is given by
F $= \frac{G M}{{(3 a + a)}^{2}} + 0 = \frac{G M}{16 a^{2}}$

At point P₂, the gravitational field due to the sphere and the shell is given by
$F = \frac{GM}{{(a + 4 a + a)}^{2}} + \frac{GM}{{(4 a + a)}^{2}} \Rightarrow F = \frac{GM}{36 a^{2}} + \frac{GM}{25 a^{2}} \Rightarrow F = \frac{GM}{a^{2}} (\frac{1}{36} + \frac{1}{25}) \Rightarrow F = \frac{GM}{a^{2}} (\frac{25 + 36}{900}) \Rightarrow F = (\frac{61}{900}) \frac{GM}{a^{2}}$

Page No 226:

Answer:

We know that in a thin spherical shell of uniform density, the gravitational field at its internal point is zero. So, at points A and B, the gravitational fields are equal and opposite and, thus, cancel each other. So the net field is zero.
Hence, E_A = E_B

Page No 226:

Answer:

Let the mass of 0.10 kg be at a distance x from the 2 kg mass and at a distance of (2 − x) from the 4 kg mass.
Force between 0.1 kg mass and 2 kg mass = force between 0.1 kg mass and 4 kg mass

$∴ \frac{2 \times 0.1}{x^{2}} = - \frac{4 \times 0.1}{{(2 - x)}^{2}} \Rightarrow \frac{0.2}{x^{2}} = \frac{0.4}{{(2 - x)}^{2}} \Rightarrow \frac{1}{x^{2}} = \frac{2}{{(2 - x)}^{2}} \Rightarrow {(2 - x)}^{2} = 2 x^{2} \Rightarrow 2 - x = \sqrt{2} x \Rightarrow x (\sqrt{2} + 1) = 2 \Rightarrow x = \frac{2}{2.414} = 0.83 m from the 2 kg mass$

Now, gravitation potential energy of the system is given by
$U = - G [\frac{0.1 \times 2}{0.83} + \frac{0.1 \times 4}{1.17} + \frac{2 \times 4}{2}] \Rightarrow U = - 6.67 \times 10^{11} [\frac{0.1 \times 2}{0.83} + \frac{0.1 \times 4}{1.17} + \frac{2 \times 4}{2}] \Rightarrow U = - 3.06 \times 10^{- 10} J$

Page No 226:

Answer:

The work done in increasing the side of the triangle from a to 2a is equal to the difference of the potential energies of the system.
i.e., work done = final potential energy of the system $-$ initial potential energy of the system

$∴ W = - 3 \frac{G m^{2}}{2 a} - (- 3 \frac{G m^{2}}{a}) \Rightarrow W = 3 \frac{G m^{2}}{2 a}$

Page No 226:

Answer:

The work done against the gravitational force to take the particle away from the sphere to infinity is equal to
the difference between the potential energy of the particle at infinity and potential energy of the particle at the surface of the sphere.

$∴ W = 0 - (- \frac{G \times 10 \times 0.1}{1 \times 0.1}) = \frac{6.67 \times 10^{- 11} \times 1}{1 \times 0.1} = 6.67 \times 10^{- 10} J$

Page No 226:

Answer:

Gravitational field, $\vec{E} = (5 N / kg) \hat{i} + (12 N / kg) \hat{j}$
(a) $\vec{F} = m \vec{E}$
$= 2 kg [(5 N / kg) \hat{i} + (12 N / Kg) \hat{j}] = (10 N) \hat{i} + (24 N) \hat{j} ∴ |\vec{F}| = \sqrt{100 + 576} = \sqrt{676} = 26 N$

(b) $V = - \vec{E .} \vec{r}$
Potential at (12 m, 0) $= - 60 J / Kg$

Potential at (0, 5 m) = $- 60 J / kg$

(c)
change in potential=final potential -initial potential
initial potential=potential at the origin=0
final potential=potential at (12,5)
$V = - \vec{E} . \vec{r} = - (10 \hat{i} + 24 \overset{⏜}{j}) . (12 \overset{⏜}{i} + 5 \overset{⏜}{j}) = - (120 + 120) J = - 240 J$

(d)
$∆ V = - \vec{E} . ∆ \vec{r} ∆ \vec{r} = (12 \overset{⏜}{i} + 0 \overset{⏜}{j}) - (0 \overset{⏜}{i} + 5 \overset{⏜}{j}) = 12 \overset{⏜}{i} - 5 \overset{⏜}{j} ∆ V = - (10 \overset{⏜}{i} + 12 \overset{⏜}{j}) . (12 \overset{⏜}{i} - 5 \overset{⏜}{j}) = 0 J$

Page No 227:

Answer:

(a) $V = (\frac{20 N}{kg}) (x + y)$
$[\frac{GM}{R}] = [\frac{{MLT}^{- 2}}{M}] [L] \Rightarrow [\frac{M^{- 1} L^{3} T^{- 2} M^{1}}{L}] = [\frac{{ML}^{2} T^{- 2}}{M}] \Rightarrow [M^{0} L^{2} T^{- 2}] = [M^{0} L^{2} T^{- 2}] ∴ LHS = RHS$

(b) The gravitational field at the point (x, y) is given by ${\vec{E}}_{(x, y)} = - 20 (\frac{N}{kg}) \hat{i} - (\frac{20 N}{kg}) \hat{j}$ .

(c) $\vec{F} = \vec{E} m$
$= 0.5 kg [- (\frac{20 N}{kg}) \hat{i} - (\frac{20 N}{kg}) \hat{j}] = - (10 N) \hat{i} - (10 N) \hat{j}$
$∴ |\vec{F}| = \sqrt{(100) + (100)} = 10 \sqrt{2} N$

Page No 227:

Answer:

The gravitational field in a region is given by $\vec{E} = 2 \hat{i} + 3 \hat{j}$ .
Slope of the electric field, $m_{1} = \tan θ_{1} = \frac{3}{2}$
The given line is 3y + 2x = 5.
Slope of the line, $m_{2} = \tan θ_{2} = - \frac{2}{3}$
We can see that m₁m₂ = −1

Since the directions of the field and the displacement are perpendicular to earth other, no work is done by the gravitational field when a particle is moved on the given line.

Page No 227:

Answer:

Assume that at height h, the weight of the body becomes half.
Weight of the body at the surface = mg
Weight of the body at height h above the Earth's surface = mg', where g' is the acceleration due to gravity at height h
$Now, g' = \frac{1}{2} g$
$∴ (\frac{1}{2}) \frac{G M}{R^{2}} = \frac{G M}{{(R + h)}^{2}} [∵ g = \frac{G M}{R^{2}}] \Rightarrow 2 R^{2} = {(R + h)}^{2} \Rightarrow \sqrt{2} R = R + h \Rightarrow h = (\sqrt{2} - 1) R$

Page No 227:

Answer:

Let g' be the acceleration due to gravity on Mount Everest.
$Then g' = g (1 - \frac{2 h}{R}) here h = 8848 m = 9.8 (1 - 0.00276) = 9.0 (0.99724) = 9.77 m / s^{2}$

∴ The acceleration due to gravity on the top of Mount Everest is 9.77 m/s².

Page No 227:

Answer:

Let g' be the acceleration due to gravity in a mine of depth d.
$∴ g' = g (1 - \frac{d}{R})$
$= 9.8 (1 - \frac{640}{640 \times 10^{3}}) = 9.8 (\frac{10000 - 1}{10^{4}}) = \frac{9.8}{10^{4}} \times 9999 = 9.8 \times 0.9999 = 9.799 m / s^{2}$

Page No 227:

Answer:

Let g_p be the acceleration due to gravity at the poles.
Let g_e be the acceleration due to gravity at the equator.

Now, acceleration due to gravity at the equator is given by
g_e = g_p $-$ ω²r
   = 9.81 − (7.3 × 10⁻⁵)² × 6400 × 10³
   = 9.81 − (53.29 × 10⁻¹⁰) × 64 × 10⁵
   = 9.81 − 0.034 = 9.776 m/s²
Now, mg_e = 1 kg × 9.776 m/s²
            = 9.776 N
∴ The body will weigh 9.776 N at the equator.

Page No 227:

Answer:

At the equator, g' = g − ω²R ...(i)
Let h be the height above the South Pole where the body stretch the spring by the same length.
The acceleration due to gravity at this point is $g' = g (1 - \frac{2 h}{R})$ .

Weight of the body at the equator = weight of the body at height h above the South Pole
$∴ g - ω^{2} r = g (1 - (\frac{2 h}{R})) \Rightarrow 1 - \frac{ω^{2} R}{2 g} = 1 - \frac{2 h}{R} \Rightarrow h = \frac{ω^{2} R^{2}}{2 g} = \frac{{(7.3 \times 10^{- 5})}^{2} \times {(6400 \times 10^{- 3})}^{2}}{2 \times 9.81} = \frac{{(7.3)}^{2} \times {(64)}^{2}}{19.62} = 11125 m = 10 km (approx .)$

Page No 227:

Answer:

The apparent acceleration due to gravity at the equator becomes zero.
i.e., g' = g − ω²R = 0
⇒ g = ω²R
$\Rightarrow ω = \sqrt{\frac{g}{R}} = \sqrt{\frac{9.8}{6400 \times 10^{3}}} \Rightarrow ω = \sqrt{\frac{9.8 \times 10^{- 5}}{6.4}} = \sqrt{1.5 \times 10^{- 6}} \Rightarrow ω = 1.2 \times 10^{- 3} rad / s ∴ T = \frac{2 π}{ω} = \frac{2 \times 3.14}{1.2 \times 10^{- 3}} = \frac{6.28}{1.2 \times 10^{- 3}} = 1.41 h$

Page No 227:

Answer:

(a) Speed of the ship due to rotation of the Earth is v = ωR, where R is the radius of the Earth and ω is its angular speed.

(b) The tension in the string is given by
T₀ = mg − mω²R
∴ T₀ − mg = mω²R

(c) Let the ship move with a speed v.
Then the tension in the string is given by

$T = m g - m ω_{1}^{2} R = T_{0} - \frac{{(v - ω R)}^{2}}{R^{2}} R = T_{0} - \frac{(v^{2} + ω^{2} R^{2} - 2 ω R v)}{R} R ∴ T = T_{0} + 2 ω v m$

Page No 227:

Answer:

According to Kepler's laws of planetary motion, the time period of revolution of a planet about the Sun is directly proportional to the cube of the distance between their centres.
i.e., T² $\propto$ R³
$\Rightarrow \frac{T_{m}^{2}}{T_{e}^{2}} = \frac{R_{m}^{3}}{R_{e}^{3}} \Rightarrow {(\frac{R_{m}}{R_{e}})}^{3} = {(\frac{1.88}{1})}^{2} ∴ \frac{R_{m}}{R_{e}} = {(1.88)}^{2 / 3} = 1.52$

Page No 227:

Answer:

Time period of rotation of the Moon around the Earth is given by
$T = 2 π \sqrt{\frac{r^{3}}{G M}}$ ,
where r is the distance between the centres of the Earth and Moon and m is the mass of the Earth.

$Now, 27.3 = 2 \times 3.14 \sqrt{\frac{{(3.84 \times 10^{5})}^{3}}{6.67 \times 10^{- 11} M}} \Rightarrow 27.3 \times 27.3 = \frac{2 \times 3.14 \times {(3.84 \times 10^{5})}^{3}}{6.67 \times 10^{- 11} M} \Rightarrow M = \frac{2 \times {(3.14)}^{2} \times {(3.84)}^{3} \times 10^{15}}{3.335 \times 10^{- 11} \times {(27.3)}^{2}} = 6.02 \times 10^{24} kg$
∴ The mass of the Earth is found to be 6.02 × 10²⁴ kg.

Page No 227:

Answer:

Time period of revolution of the satellite around the Mars is give by
$T = 2 π \sqrt{\frac{r^{3}}{G M}}$ ,
where M is the mass of the Mars and r is the distance of the satellite from the centre of the planet.

$Now, 27540 = 2 \times 3.14 \sqrt{\frac{{(9.4 \times 10^{3} \times 10^{3})}^{3}}{6.67 \times 10^{- 11} \times M}} \Rightarrow {(27540)}^{2} = {(6.28)}^{2} \times \frac{{(9.4 \times 10^{5})}^{3}}{6.67 \times 10^{- 11} \times M} \Rightarrow M = \frac{{(6.28)}^{2} \times {(9.4)}^{3} \times 10^{18}}{6.67 \times 10^{- 11} \times {(27540)}^{2}} \Rightarrow M = 6.5 \times 10^{23} kg$

Page No 227:

Answer:

(a) Speed of the satellite in its orbit
$v = \sqrt{\frac{G M}{r + h}} = \sqrt{\frac{g r^{2}}{r + h}}$
$\Rightarrow v = \sqrt{\frac{9.8 \times {(6400 \times 10^{3})}^{2}}{10^{6} \times (6.4 + 2)}} \Rightarrow v = \sqrt{\frac{9.8 \times 6.4 \times 6.4 \times 10^{6}}{8.4}} \Rightarrow v = 6.9 \times 10^{3} m / s = 6.9 km / s$

(b) Kinetic energy of the satellite
$K . E . = \frac{1}{2} m v^{2}$
$= \frac{1}{2} \times 1000 \times {(6.9 \times 10^{3})}^{2} = \frac{1}{2} \times 1000 \times (47.6 \times 10^{6}) = 2.38 \times 10^{10} J$

(c) Potential energy of the satellite
$P . E . = - \frac{GM m}{(R + h)}$
$= - \frac{6.67 \times 10^{- 11} \times 6 \times 10^{24} \times 10^{3}}{(6400 + 2000) \times 10^{3}} = \frac{40 \times 10^{13}}{8400} = - 4.76 \times 10^{10} J$

(d) Time period of the satellite
$T = \frac{2 π (r + h)}{v}$
$= \frac{2 \times 3.14 \times 8400 \times 10^{3}}{6.9 \times 10^{3}} = \frac{6.28 \times 84 \times 10^{2}}{6.9} = 76.6 \times 10.2 s = 2.1 h$

Page No 227:

Answer:

(a) The angular speed of the Earth and the satellite will be the same.

$i . e ., \frac{2 π}{T_{e}} = \frac{2 π}{T_{s}} \Rightarrow \frac{1}{24 \times 3600} = \frac{1}{2 π \sqrt{{(R + h)}^{3} / g h^{2}}} \Rightarrow 12 \times 3600 = 3.14 \sqrt{\frac{{(R + h)}^{3}}{g R^{2}}} \Rightarrow \frac{{(R + h)}^{3}}{g R^{2}} = \frac{{(12 \times 3600)}^{2}}{{(3.14)}^{2}} \Rightarrow \frac{(6400 + h^{3}) \times 10^{9}}{9.8 \times {(6400)}^{2} \times 10^{6}} = \frac{{(12 \times 3600)}^{2}}{{(3.14)}^{2}} \Rightarrow \frac{(6400 + h) \times 10^{9}}{6272 \times 10^{9}} = 432 \times 10^{4} \Rightarrow {(6400 + h)}^{3} = 6272 \times 432 \times 10^{4} \Rightarrow 6400 + h = {(6272 \times 432 \times 10^{4})}^{1 / 3} - 6400 \Rightarrow h = 42300 km$

(b) Time taken from the North Pole to the equatorial plane is given by
$\frac{1}{4} T = \frac{1}{4} \times 6.28 \sqrt{\frac{{(42300 + 6400)}^{3}}{10 \times {(6400)}^{2} \times 10^{6}}} = 3.14 \sqrt{\frac{{(479)}^{3} \times 10^{6}}{{(64)}^{2} \times 10^{11}}} = 3.14 \sqrt{\frac{497 \times 497 \times 497}{64 \times 64 \times 10^{5}}} = 6 h$

Page No 227:

Answer:

For a geostationary satellite, we have:
R = 6.4 × 10³ km
h = 3.6 × 10³ km
Given: mg = 10 N

The true weight of the object in the geostationary satellite is given by
$m g' = m g - \frac{R^{2}}{{(R + h)}^{2}} = 10 - \frac{{(6400 \times 10^{3})}^{2}}{(6400 \times 10^{3} + 3600 \times 10^{3})} = 10 - [\frac{{(64 \times 10^{5})}^{2}}{(6.4 \times 10^{6} + 36 \times 10^{5})}] = 10 - [\frac{4096 \times 10^{10}}{{(42.4)}^{2} \times 10^{12}}] = \frac{4096}{17980} = 0.227 N$

Page No 227:

Answer:

The time period of revolution of the satellite around a planet in terms of the radius of the planet and radius of the orbit of the satellite is given by $T = 2 π \sqrt{\frac{R_{2}^{2}}{g R_{1}^{2}}}$ , where g is the acceleration due to gravity at the surface of the planet.
$Now, T^{2} = 4 π^{2} \frac{R_{2}^{2}}{g R_{1}^{2}} \Rightarrow g = \frac{4 π^{2}}{T^{2}} \frac{R_{2}^{2}}{R_{1}^{2}}$
∴ Acceleration due to gravity of the planet = $\frac{4 π^{2}}{T^{2}} \frac{R_{2}^{2}}{R_{1}^{2}}$

Page No 227:

Answer:

Consider that B is the position of the geostationary satellite.

In the given figure, $ϕ$ is the latitude and θ is the colatitude of a place which can directly receive a signal from a geostationary satellite.

In triangle OAB, we have:
$\cos ϕ = \frac{6400}{42000} = \frac{16}{106} = \frac{8}{53} i . e ., ϕ = \cos^{- 1} \frac{8}{53} = \cos^{- 1} 0.15 Now, θ = \frac{π}{2} - ϕ \Rightarrow θ = \frac{π}{2} - \cos^{- 1} 0.15 \Rightarrow θ = \sin^{- 1} 0.15$

Page No 227:

Answer:

The particle attained a maximum height 6400 km, which is equal to the radius of the Earth.

Total energy of the particle on the Earth's surface is given by
$E_{e} = \frac{1}{2} M V^{2} + (\frac{- g m M}{r}) . . . . (1)$

Now, total energy of the particle at the maximum height is given by
$E_{3} = (\frac{- G M m}{R + h}) + 0 \Rightarrow E_{3} = (\frac{- G M m}{2 R}) . . . (2) (∴ g = R)$
From equations (1) and (2), we have:

$- \frac{G M m}{R} + \frac{1}{2} m v^{2} = - \frac{G M m}{2 R} \Rightarrow (\frac{1}{2}) m v^{2} = G M m (- \frac{1}{2 R} + \frac{1}{R}) \Rightarrow v^{2} = \frac{G M}{R} = \frac{6.67 \times 10^{- 11} \times 6 \times 10^{24}}{6400 \times 10^{3}} = \frac{40.02 + 10^{13}}{6.4 \times 10^{6}} = 6.2 \times 10^{7} = 0.62 \times 10^{8} ∴ v = \sqrt{0.62 \times 10^{8}} = 0.79 \times 10^{4} m / s = 79 km / s$

Page No 227:

Answer:

Initial velocity of the particle, v = 15 km/s
Let its speed be v' in interstellar space.

Applying the law of conservation of energy, we have:
$(\frac{1}{2}) m [v - v'^{2}] = \int_{R}^{\infty} \frac{G M m}{x^{2}} d x ∴ (\frac{1}{2}) m [15 \times 10^{3} - v'^{2}] = \int_{R}^{\infty} \frac{G M m}{x^{2}} d x \Rightarrow (\frac{1}{2}) m [{(15 \times 10^{3})}^{2} - v'^{2}] = GM m [\frac{- 1}{x}] \Rightarrow (\frac{1}{2}) m [(225 \times 10^{5}) - v'^{2}] = \frac{GM m}{R} \Rightarrow 225 \times 10^{5} - v'^{2} = \frac{2 \times 6.67 \times 10^{- 11} \times 6 \times 10^{24}}{6400 \times 10^{3}} \Rightarrow v'^{2} = 225 \times 10^{6} - \frac{40.02}{32} \times 10^{8} = 2.25 \times 10^{8} - 1.2 \times 10^{8} = 10^{8} (1.05) Or v' = 1.01 \times 10^{4} m / s = 10 km / s$

Page No 227:

Answer:

Mass of the sphere = 6 × 10²⁴ kg
Escape velocity = 3 × 10⁸ m/s

Escape velocity is given by
$v_{e} = \frac{2 G M}{R} \Rightarrow R = \frac{2 G M}{{v_{e}}^{2}} = \frac{2 \times 6.67 \times 10^{- 11} \times 6 \times 10^{24}}{{(3 \times 10^{8})}^{2}} = \frac{2 \times 40.02 \times 10^{13}}{9 \times 10^{16}} = \frac{80.02}{9} \times 10^{- 3} m = 8.89 \times 10^{- 3} m = 9 mm$

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