Ncert Solution CBSE Class 11 Physics Chapter 7 Gravitation - Param Himalaya
7.1 Answer the following :
(a) You can shield a charge from electrical forces by putting it inside a hollow conductor.Can you shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means ?
Solution: You cannot shield a body from the gravitational influence of nearby matter by any means. It is because the gravitational force on a body due to nearby matter is not altered due to the presence of other bodies. In other words , gravitational field cannot be shielded by any means.
(b) An astronaut inside a small space ship orbiting around the earth cannot detect gravity. If the space station orbiting around the earth has a large size, can he hope to detect gravity ?
Solution: Yes , If the size of the spaceship orbiting earth is very large , an astronaut in it can detect the gravity.
(c) If you compare the gravitational force on the earth due to the sun to that due to the moon, you would find that the Sun’s pull is greater than the moon’s pull.(you can check this yourself using the data available in the succeeding exercises). However, the tidal effect of the moon’s pull is greater than the tidal effect of sun.Why ?
Solution : The tidal effect varies inversely as the cube of distance and is not governed by inverse square law. Since sun is very far off from earth compared to the moon, the tidal effect of moon's pull is greater than the tidal effect of sun.
7.2 Choose the correct alternative :
(a) Acceleration due to gravity increases/decreases with increasing altitude.
(b) Acceleration due to gravity increases/decreases with increasing depth (assume the earth to be a sphere of uniform density).
(c) Acceleration due to gravity is independent of mass of the earth/mass of the body.
(d) The formula –G Mm(1/r2 – 1/r1) is more/less accurate than the formula mg(r2 – r1) for the difference of potential energy between two points r2 and r1 distance away from the centre of the earth.
Solution:
(a) Decrease
(b) decrease
(c) Mass of the body
(d) More
7.3 Suppose there existed a planet that went around the Sun twice as fast as the earth.What would be its orbital size as compared to that of the earth ?
Solution:
Let $r_{e}$ and $r_{p}$ be the radii of orbit of earth and planet respectively.
Now , $T_{e}$ = 1 year ;
$T_{p} = T_{e}/2 = 1/2 = 0.5 year$
According to Kepler's third law , we have ,
$$(\frac{r_{p}}{r_{e}})^{3} = (\frac{T_{p}}{T_{e}})^{2}$$
$$r_{p} = r_{e} (\frac{T_{p}}{T_{e}})^{2/3}$$
$$r_{p} = r_{e} \times (\frac{0.5}{1})^{2/3}$$
$$r_{p} = 0.63 r_{e}$$
7.4 $I_{o}$, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is $4.22 \times 10^{8}m$. Show that the mass of Jupiter is about one-thousandth that of the sun.
Solution:
When a satellite revolves around a planet of mass M in an orbit of radius r , the time period of the satellite is given by :
$$T = 2 \pi \sqrt{\frac{r^{3}}{GM}}$$
$$M = \frac{4 \pi^{2}r{3}}{GT^{2}}$$
Mass of Jupiter $M_{J}$ : The time period of satellite of Jupiter is $T_{1}$ = 1.769 days and its orbit radius is $r_{1}= 4.22 \times 10^{8}m$.
$$M_{J} = \frac{4 \pi \times (4.22 \times 10^{8})^{3}}{G \times (1.769)^{2}}$$
$$M_{J}= \frac{4 \pi^{2}}{G} \times 2.4 \times 10^{25}$$
Mass of Sun $M_{s}$ : The time period of earth around the sun is $T_{2}$= 365 days and its orbit radius
$$r_{2} = 1 A.U. = 1.496 \times 10^{11}m.$$
$$M_{S} = \frac{4 \pi \times (1.496 \times 10^{11})^{3}}{G \times (365)^{2}}$$
$$M_{S}= \frac{4 \pi^{2}}{G} \times 2.5 \times 10^{28}$$
$$\frac{M_{J}}{M_{S}} = \frac{2.4 \times 10^{25}}{2.5 \times 10^{28}}$$
$$\frac{M_{J}}{M_{S}} = \frac{1}{1000}$$
7.5 Let us assume that our galaxy consists of $2.5 \times 10^{11}$ stars each of one solar mass. How long will a star at a distance of 50,000 ly from the galactic centre take to complete one revolution ? Take the diameter of the Milky Way to be $10^{5}$ly.
Solution:
We know : 1 star mass = 1 solar mass : $2 \times 10^{30}Kg$
Mass of galaxy ,
$$M = 2.5 \times 10^{11} stars$$
$$M= 2.5 \times 10^{11} \times (2 \times 10^{30})$$
$$M= 5 \times 10^{41}kg$$
Radius of orbit of star ,
$$r = 50000 ly = 50000 \times 9.46 \times 10^{15}m$$
$$r= 4.73 \times 10^{20}m$$
Time period of star ,
$$T = 2 \pi \sqrt{\frac{r^{3}}{GM}}$$
$$T= 2 \pi \sqrt{\frac{(4.73 \times 10^{30})^{3}}{6.67 \times 10^{-11} \times 5 \times 10^{41}}}$$
$$T= 1.12 \times 10^{16}s$$
7.6 Choose the correct alternative:
(a) If the zero of potential energy is at infinity, the total energy of an orbiting satellite is negative of its kinetic/potential energy.
(b) The energy required to launch an orbiting satellite out of earth’s gravitational influence is more/less than the energy required to project a stationary object at the same height (as the satellite) out of earth’s influence.
Solution:
(a) Kinetic Energy
(b) Less
7.7 Does the escape speed of a body from the earth depend on (a) the mass of the body, (b)the location from where it is projected, (c) the direction of projection, (d) the height of the location from where the body is launched?
Solution:
(a) The escape speed of a body from earth does not depend on the mass of the body because $V_{e} = \sqrt{\frac{2GM}{R}}$.
(b) Yes , The value of g depends on latitude which is different at different locations on earth. We know that escape speed is also given by $V_{e} = \sqrt{2gR}$.
(c) It does not depend on the direction of projection.
(d) Yes , it is because $V_{e} = \sqrt{2gR}$ and the value of g depends upon the height of location from where the body is projected.
7.8 A comet orbits the sun in a highly elliptical orbit. Does the comet have a constant (a)linear speed, (b) angular speed, (c) angular momentum, (d) kinetic energy, (e) potential energy, (f) total energy throughout its orbit? Neglect any mass loss of the comet when it comes very close to the Sun.
Solution:
(a) The linear speed is not constant
(b) The angular speed is not constant
(c) The angular momentum remains constant
(d) Since the linear speed varies , the kinetic energy ($=mc^{2}$) is not constant.
(e) The potential energy is not constant because the distance of the comet from the sun changes.
(f) The total energy remains constant. Although P.E. and K.E. change but total energy (P.E + K.E.) at every point in the orbit remains the same.
7.9 Which of the following symptoms is likely to afflict an astronaut in space (a) swollen feet, (b) swollen face, (c) headache, (d) orientational problem.
Solution:
(a) On earth, the weight of the body of a person is carried by his legs. But in space the astronaut is at the state of weightlessness. Therefore, the feet of the astronaut do not get swollen.
(b) In free space, the face of the astronaut is expected to get more blood supply. Therefore, the face of the astronaut may get swollen.
(c) In the gravity free space, the swollen face of the astronaut may cause headache.
(d) Space has different orientations. Therefore, orientational problem can affect an astronaut in space.
7.10 In the following two exercises, choose the correct answer from among the given ones:
The gravitational intensity at the centre of a hemispherical shell of uniform mass density has the direction indicated by the arrow (see Fig 7.11) (i) a, (ii) b, (iii) c, (iv) 0.
7.11 For the above problem, the direction of the gravitational intensity at an arbitrary point P is indicated by the arrow (i) d, (ii) e, (iii) f, (iv) g.
7.12 A rocket is fired from the earth towards the sun. At what distance from the earth’s centre is the gravitational force on the rocket zero ? Mass of the sun =$ 2 \times 10^{30} kg$, mass of the earth = $6 \times 10^{24} kg$. Neglect the effect of other planets etc. (orbital radius= $1.5 \times 10^{11} m$).
Solution:
Suppose the gravitational forces on the rocket due to the sun and earth become equal at a distance x from the centre of earth. If m is the mass of rocket , then ,
$$\frac{GM_{S}m}{(r-x)^{2}} = \frac{GM_{E}m}{x^{2}}$$
$$\frac{(r-x)^{2}}{x^{2}} = \frac{M_{S}}{M_{E}}$$
$$\frac{r-x}{x} = \sqrt{\frac{M_{S}}{M_{E}}}$$
$$\frac{r-x}{x} = \sqrt{\frac{2 \times 10^{30}}{6 \times 10^{24}}}$$
$$\frac{r-x}{x} = 577.35$$
$$\frac{r}{x}-1 = 577.35$$
$$\frac{r}{x} = 578.35$$
$$x = \frac{r}{578.35}$$
$$x = \frac{1.5 \times 10^{11}}{578.35}$$
$$x= 2.59 \times 10^{8}m$$
7.13 How will you ‘weigh the sun’, that is estimate its mass? The mean orbital radius of the earth around the sun is $1.5 × 10^{8}km$
Solution:
Radius of Earth's orbit = R+h = $1.5 \times 10^{8} km$ = $1.5 \times 10^{11}m$
Period of earth around sun , T = 365 days = 365 × 24 × 60 × 60 sec
The time period is given by ;
$$T = 2 \pi \sqrt{\frac{(R+h)^{3}}{GM}}$$
Or Mass of sun ,
$$M = \frac{4 \pi^{2}(R+h)^{3}}{GT^{2}}$$
$$M = \frac{ 4 \pi^{2} \times (1.5 \times 10^{11})^{3}}{(6.67 \times 10^{-11}) \times (365 \times 24 \times 60 \times 60)^{2}}$$
$$M = 2.01 \times 10^{30} kg$$
7.14 A saturn year is 29.5 times the earth year. How far is the saturn from the sun if the earth is $1.50 × 10^{8}$ km away from the sun ?
Solution:
Time period of earth , $T_{1}$ = 1 year$
Time period of Saturn, $T_{2}$ = 29.5 years
Distance of each from sun , $r_{1} = 1.5 \times 10^{8}m$
If $r_{2}$ is the distance of the Saturn from sun , then according to Kepler's third law,
$$\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{r_{1}^{3}}{r_{2}^{3}}$$
$$r_{2} = r_{1}(\frac{T_{2}^{2}}{T_{1}^{2}})^{2/3}$$
$$r_{2} = 1.5 \times 10^{8} \times (\frac{29.5}{1})^{2/3}$$
$$r_{2} = 14.32 \times 10^{11}m$$
7.15 A body weighs 63 N on the surface of the earth. What is the gravitational force on it due to the earth at a height equal to half the radius of the earth ?
Solution:
Weight of body = mg = 63 N ; h = R/2
The value of acceleration due to gravity ($g^{'}$) at height h is
$$g^{'}= \frac{gR^{2}}{(R+h)^{2}}$$
$$g^{'}= \frac{gR^{2}}{(R+\frac{R}{2})^{2}}$$
$$g^{'}= \frac{gR^{2}}{R^{2}(1+\frac{1}{2})^{2}}$$
$$g^{'}= \frac{g}{(\frac{3}{2})^{2}}$$
$$g^{'}= \frac{g}{\frac{9}{4}}$$
$$g^{'}= \frac{4}{9}g$$
$\therefore$ Gravitational force on the body at height h is
$$F=mg^{'}$$
$$F= m \times \frac{4}{9}g$$
$$F = \frac{4}{9} \times mg$$
$$F = \frac{4}{9} \times 63$$
$$F = 28 N$$
7.16 Assuming the earth to be a sphere of uniform mass density, how much would a body weigh half way down to the centre of the earth if it weighed 250 N on the surface ?
Solution:
Weight of a boby of mass m at the earth's surface,
W = mg
W= 250 N
Boby of mass m is located at depth ,$d = \frac{R_{e}}{2}$
Acceleration due to gravity at depth g(d) is given by the relation:
$$g^{'} = g (1-\frac{d}{R})$$
$$mg^{'} = mg (1-\frac{d}{R})$$
$$W^{'} = 250(1-\frac{R/2}{R})$$
$$W^{'} = 250(1-\frac{1}{2})$$
$$W^{'} = 250 \times \frac{1}{2}$$
$$W^{'} = 125 N$$
7.17 A rocket is fired vertically with a speed of 5 km/s from the earth’s surface. How far from the earth does the rocket go before returning to the earth ? Mass of the earth= $6 \times 10^{24}$ kg mean radius of the earth = $6.4 \times 10^{6}$ m ; G = $6.67 \times 10^{-11} Nm^{2}kg^{-2}$
Solution:
Here v = 5 Km/s = 5000 m/s ;
$M = 6 \times 10^{24}kg$
$R = 6.4 \times 10^{6}m$
Suppose the rocket goes to a height h before returning to the earth. Clearly , at this height the kinetic energy of the rocket is zero. According to principle of conservation of energy,
$$K.E + P.E ( At \ earth ) = K.E + P.E ( At \ height \ h )$$
$$\frac{1}{2}mv^{2} - \frac{GMm}{R} =0 + (-\frac{GMm}{(R+h)}$$
$$v^{2} - \frac{2GM}{R} = -\frac{2GM}{R+h}$$
$$\frac{2GM}{R+h} = \frac{2GM}{R}-v^{2}$$
$$\frac{2GM}{R+h}= \frac{2GM-v^{2}R}{R}$$
$$R+h = \frac{2GMR}{2GM-v^{2}R}$$
$R+h= \frac{2 \times (6.67 \times 10^{-11}) \times (6 \times 10^{24}) \times 6.4 \times 10^{6}}{2 \times 6.67 \times 10^{-11} \times 6 \times 10^{24} - (5000)^{2} \times 6.4 \times 10^{6}}$
$$R+h= 8 \times 10^{6}m$$
$$h = 8 \times 10^{6} - R$$
$$h = 8 \times 10^{6} -6.4 \times 10^{6}$$
$$h = 1.6 \times 10^{6}m$$
7.18 The escape speed of a projectile on the earth’s surface is 11.2 km s–1. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets.
Solution:
Escape velocity, $V_{e} = 11.2 km/s = 11.2 \times 10^{3} m/s$
Velocity of projection of projectile , $v= 3 V_{e}$
Suppose M is the mass of the projectile and $v_{0}$ is it's velocity after escaping the gravitational pull.
Applying the law of conservation of energy , we have ,
$$\frac{1}{2}Mv_{0}^{2} = \frac{1}{2}Mv^{2}+ \frac{1}{2}MV_{e}^{2}$$
$$v_{0}^{2} = v^{2} - V_{e}^{2}$$
$$v_{0}^{2} = (3V_{e})^{2} - V_{e}^{2}$$
$$v_{0}^{2} = 8V_{e}^{2}$$
$$v_{0} = \sqrt{8v_{e}^{2}}$$
$$v_{0} = v_{e}\sqrt{8}$$
$$v_{0} = 11.2 \times 10^{3}\sqrt{8}
$$v_{0} = 31.68 \times 10^{3} m/s
$$v_{0} = 31.68 km/s$$
7.19 A satellite orbits the earth at a height of 400 km above the surface. How much energy must be expended to rocket the satellite out of the earth’s gravitational influence? Mass of the satellite = 200 kg; mass of the earth = $6.0 \times 10^{24} kg$: radius of the earth = $6.4 \times 10^{6}m$; G = $6.67 \times 10^{11}N m^{2}kg^{–2}$.
Solution:
Here $G = 6.67 \times 10^{-11} Nm^{2}kg^{-2}$
$M = 6.0 \times 10 ^{24} kg$
m= 200 kg
$R = 6.4 \times 10^{6}m$
$h = 400km = 400 \times 10^{3} = 4 \times 10^{5} m$
The K.E and P.E of a satellite at a height h above the surface of earth are :
$$K.E \ of \ Satellite = \frac{1}{2}mv^{2} = \frac{GMm}{2(R+h)}$$
$$P.E \ of \ Satellite = -\frac{GMm}{R+h}$$
$$Total \ Energy = K.E + P.E$$
$$E= \frac{GMm}{2(R+h)} -\frac{GMm}{R+h}$$
$$E= -\frac{GMm}{2(R+h)}$$
$$E = -\frac{(6.67 \times 10^{11}) \times (6.0 \times 10^{24}) \times 200}{2(6.4 \times 10^{6}+4 \times 10^{5})}$$
$$E =- \frac{6.67 \times 6.0 \times 10^{15}}{6.8 \times 10^{6}}$$
$$E = -5.89 \times 10^{9} J$$
The negative sign indicates that the satellite is bound to the Earth.
The total energy of the Satellite at infinity is zero. Therefore, the energy required for the satellite to leave its orbit around the earth and escape to infinity
Energy required = 0 - E
= $0-(-5.89 \times 10^{9} J)$
Energy required = $5.89 \times 10^{9} J$
This is also the binding energy of the satellite.
7.20 Two stars each of one solar mass ($2 \times 10^{30}$kg) are approaching each other for a head on collision. When they are a distance $10^{9} km$, their speeds are negligible.What is the speed with which they collide ? The radius of each star is $10^{4} km$. Assume the stars to remain undistorted until they collide. (Use the known value of ($G= 6.67 \times 10^{-11} Nm^{2} kg^{-2}$)
Solution:
Mass of each star , $M = 2 \times 10^{30} kg$ ;
Radius of each star , $R = 10^{4} km = 10^{7} m$
Distance between the stars , $r=10^{9}km (=10^{12}m)$
For negligible speeds , v = 0 total energy of two stars separated at distance r
$$=- \frac{GM.M}{r}+\frac{1}{2}mv^{2}$$
$$=- \frac{GM.M}{r}+ 0 ....(i)$$
Now , consider the case when the stars are about to collide :
Velocity of the stars = v
Distance between the center's of the stars = 2R
Total kinetic energy of both stars $= \frac{1}{2}mv^{2} +\frac{1}{2}mv^{2} = Mv^{2}$
Total potential energy of both stars $= - \frac{GMM}{2R}$
Total energy of two stars = $Mv^{2} - \frac{GMM}{2R}$. .... (ii)
Using the law of conservation of energy , we can write :
$$Mv^{2}- \frac{GMM}{2R} = \frac{-GMM}{r}$$
$$v^{2} = -\frac{GM}{r}+\frac{GM}{2R} = GM (-\frac{1}{r}+\frac{1}{2R})$$
$$= 6.67 \times 10 ^{-11} \times 2 \times 10^{30}[-\frac{1}{10^{12}}+\frac{1}{2 \times 10^{7}}]$$
$$= 13.34 \times 10^{19}[-10^{-12}+5\times10^{-8}]$$
$$\sim 13.34 \times 10^{19} \times 5 \times 10^{-8}$$
$$\sim 6.67 \times 10^{12}$$
$$v= \sqrt{6.67 \times 10^{12}} $$
$$v= 2.58 \times 10^{6} m/s$$
7.21 Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the mid point of the line joining the centres of the spheres ? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable ?
Solution:
Here CA = CB = 0.5 m = r
Force of gravitational field at C due to sphere A
$$F_{CA} = \frac{Gm}{(0.5)^{2}} Along CA$$
Force of gravitational field at C due to sphere B
$$F_{CB} = \frac{Gm}{(0.5)^{2}} Along CB$$
$$F_{Net} = \frac{Gm}{(0.5)^{2}} - \frac{Gm}{(0.5)^{2}} $$
$$F_{Net} = 0$$
Since the gravitational forces at C due to the two spheres are equal in magnitude and opposite in direction, the net gravitational force at C is zero.
Gravitational potential at C due to the two spheres is
$$V_{C} = (-\frac{Gm}{r})+(-\frac{Gm}{r})$$
$$V_{C} = -\frac{2Gm}{r}$$
$$V_{C} = -\frac{2 \times (6.67 \times 10^{-11}) \times 100}{0.5}$$
$$V_{C} = -2.7 \times 10^{-8} J kg^{-1}$$
Since the net force on the object placed at the mid point C is zero, the object is in equilibrium. If the object is displaced from mid-point C towards either sphere , the objects will not return to its initial position of equilibrium. Therefore, the objects is in unstable equilibrium.