Chapter 8 Gravitation Solutions
Question - 11 : - Choose the correct answerfrom among the given ones:
Forthe problem 8.10, the direction of the gravitational intensity at anarbitrary point P is indicated by the arrow (i) d, (ii) e, (iii) f, (iv) g.
Answer - 11 : -
Answer: (ii)
Gravitationalpotential (V) is constant at all points in a spherical shell. Hence, thegravitational potential gradient 
is zero everywhere inside the sphericalshell. The gravitational potential gradient is equal to the negative ofgravitational intensity. Hence, intensity is also zero at all points inside thespherical shell. This indicates that gravitational forces acting at a point ina spherical shell are symmetric.If the upper half of aspherical shell is cut out (as shown in the given figure), then the netgravitational force acting on a particle at an arbitrary point P will be in thedownward direction.
Sincegravitational intensity at a point is defined as the gravitational force perunit mass at that point, it will also act in the downward direction. Thus, thegravitational intensity at an arbitrary point P of the hemispherical shell hasthe direction as indicated by arrow e.
Question - 12 : - A rocket is fired from the earth towards thesun. At what distance from the earth’s centre is the gravitational force on therocket zero? Mass of the sun = 2 ×1030 kg, mass of theearth = 6 × 1024 kg. Neglect the effect of otherplanets etc. (orbital radius = 1.5 × 1011 m).
Answer - 12 : -
Mass of the Sun, Ms = 2 × 1030 kg
Mass of the Earth, Me = 6 × 10 24 kg
Orbital radius, r = 1.5 ×1011 m
Mass of the rocket = m
Let x be the distance fromthe centre of the Earth where the gravitational force acting on satellite Pbecomes zero.
From Newton’s law ofgravitation, we can equate gravitational forces acting on satellite P under theinfluence of the Sun and the Earth as:
Question - 13 : - How will you ‘weigh the sun’, that isestimate its mass? The mean orbital radius of the earth around the sun is 1.5 ×108 km.
Answer - 13 : -
Orbital radius of the Earth around theSun, r = 1.5 × 1011 m
Time taken by the Earth tocomplete one revolution around the Sun,
T = 1 year = 365.25 days
= 365.25 × 24 × 60 × 60 s
Universal gravitational constant, G = 6.67 ×10–11 Nm2 kg–2
Thus, mass of the Sun canbe calculated using the relation,
Hence,the mass of the Sun is 2 × 1030 kg.
Question - 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 ×108 km away from thesun?
Answer - 14 : -
Distance of the Earth from the Sun, re = 1.5 × 108 km = 1.5 × 1011 m
Time period of the Earth = Te
Time period of Saturn, Ts = 29. 5 Te
Distance of Saturn from the Sun = rs
From Kepler’s third law ofplanetary motion, we have
For Saturn and Sun, we canwrite
Hence, the distance between Saturn and the Sunis
Question - 15 : - Abody weighs 63 N on the surface of the earth. What is the gravitational forceon it due to the earth at a height equal to half the radius of the earth?
Answer - 15 : -
Weight of the body, W = 63N
Acceleration due to gravity at height h fromthe Earth’s surface is given by the relation:
Where,
g = Acceleration due togravity on the Earth’s surface
Re = Radius of the Earth
Weight of a body of mass m atheight h is given as:
Question - 16 : - Assumingthe earth to be a sphere of uniform mass density, how much would a body weighhalf way down to the centre of the earth if it weighed 250 N on the surface?
Answer - 16 : -
Weight of a body of mass m atthe Earth’s surface, W = mg = 250 N
Bodyof mass m is located at depth, 
Where,
= Radius of the Earth
Acceleration due to gravity at depth g (d)is given by the relation:
Weight of the body at depth d,
A rocket is fired vertically with a speed of5 km s–1 from the earth’s surface. How far fromthe earth does the rocket go before returning to the earth? Mass of the earth =6.0 × 1024 kg; mean radius of the earth = 6.4 ×106 m; G= 6.67 × 10–11 N m2 kg–2.
Answer - 17 : -
8 × 106 m from the centre ofthe Earth
Velocity of the rocket, v =5 km/s = 5 × 103 m/s
Massof the Earth,
Radius of the Earth,
Height reached by rocket mass, m = h
At the surface of theEarth,
Total energy of the rocket= Kinetic energy + Potential energy
At highest point h,
Total energy of the rocket
From the law ofconservation of energy, we have
Total energy of the rocket at the Earth’ssurface = Total energy at height h
Height achieved by therocket with respect to the centre of the Earth
The escape speed of a projectile on theearth’s surface is 11.2 km s–1. A body is projected outwith thrice this speed. What is the speed of the body far away from the earth?Ignore the presence of the sun and other planets.
Answer - 18 : -
Escape velocity of a projectile from theEarth, vesc = 11.2 km/s
Projection velocity of the projectile, vp = 3vesc
Mass of the projectile = m
Velocity of the projectile far away from theEarth = vf
Totalenergy of the projectile on the Earth 
Gravitational potentialenergy of the projectile far away from the Earth is zero.
Totalenergy of the projectile far away from the Earth = 
From the law ofconservation of energy, we have
A satellite orbits the earth at a height of400 km above the surface. How much energy must be expended to rocket thesatellite out of the earth’s gravitational influence? Mass of the satellite =200 kg; mass of the earth = 6.0 ×1024 kg; radius of theearth = 6.4 ×106 m; G = 6.67 × 10–11 N m2 kg–2.
Answer - 19 : -
Mass of the Earth, M = 6.0× 1024 kg
Mass of the satellite, m =200 kg
Radius of the Earth, Re = 6.4 × 106 m
Universal gravitational constant, G = 6.67 ×10–11 Nm2kg–2
Height of the satellite, h =400 km = 4 × 105 m = 0.4 ×106 m
Totalenergy of the satellite at height h
Orbital velocity of the satellite, v =
Total energy of height, h
The negative signindicates that the satellite is bound to the Earth. This is called bound energyof the satellite.
Energy required to sendthe satellite out of its orbit = – (Bound energy)
Two stars each of one solar mass (= 2× 1030 kg) are approachingeach other for a head on collision. When they are a distance 109 km, theirspeeds are negligible. What is the speed with which they collide? The radius ofeach star is 104 km. Assume the stars to remain undistorted until they collide.(Use the known value of G).
Answer - 20 : -
Mass of each star, M = 2 ×1030 kg
Radius of each star, R = 104 km = 107 m
Distance between the stars, r =109 km = 1012m
For negligible speeds, v =0 total energy of two stars separated at distance r
Now, consider the casewhen the stars are about to collide:
Velocity of the stars = v
Distance between the centers of the stars =2R
Totalkinetic energy of both stars 
Total potential energy of both stars 
Total energy of the two stars = 
Using the law ofconservation of energy, we can write:
