Chapter 8 Gravitation Solutions
Question - 21 : - Twoheavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on ahorizontal table. What is the gravitational force and potential at the midpoint of the line joining the centers of the spheres? Is an object placed atthat point in equilibrium? If so, is the equilibrium stable or unstable?
Answer - 21 : -
The situation isrepresented in the given figure:
Mass of each sphere, M =100 kg
Separation between the spheres, r =1m
X is the mid point betweenthe spheres. Gravitational force at point X will be zero. This is becausegravitational force exerted by each sphere will act in opposite directions.
Gravitational potential at point X:
Anyobject placed at point X will be in equilibrium state, but the equilibrium isunstable. This is because any change in the position of the object will changethe effective force in that direction.
Question - 22 : - As you have learnt in the text, ageostationary satellite orbits the earth at a height of nearly 36,000 km fromthe surface of the earth. What is the potential due to earth’s gravity at thesite of this satellite? (Take the potential energy at infinity to be zero).Mass of the earth = 6.0 × 1024 kg, radius = 6400km.
Answer - 22 : -
Mass of the Earth, M = 6.0× 1024 kg
Radius of the Earth, R =6400 km = 6.4 × 106 m
Height of a geostationarysatellite from the surface of the Earth,
h = 36000 km = 3.6 × 107 m
Gravitational potential energy due toEarth’s gravity at height h,
Question - 23 : - A star 2.5 times the mass of the sun andcollapsed to a size of 12 km rotates with a speed of 1.2 rev. per second. (Extremelycompact stars of this kind are known as neutron stars. Certain stellar objectscalled pulsars belong to this category). Will an object placed on its equatorremain stuck to its surface due to gravity? (Mass of the sun = 2 × 1030 kg).
Answer - 23 : -
Answer: Yes
A body gets stuck to thesurface of a star if the inward gravitational force is greater than the outwardcentrifugal force caused by the rotation of the star.
Gravitational force, fg
Where,
M = Mass of the star = 2.5 × 2 × 1030 = 5 × 1030 kg
m = Mass of the body
R = Radius of the star = 12 km = 1.2 ×104 m
Centrifugal force, fc = mrω2 ω =Angular speed = 2πν
ν = Angular frequency = 1.2 rev s–1
fc = mR (2πν)2
= m × (1.2 ×104) × 4 × (3.14)2 × (1.2)2 = 1.7 ×105m N
Since fg > fc, the body will remainstuck to the surface of the star.
Question - 24 : - A spaceship is stationed on Mars. How muchenergy must be expended on the spaceship to launch it out of the solar system?Mass of the space ship = 1000 kg; mass of the Sun = 2 × 1030 kg; mass of mars =6.4 × 1023 kg; radius of mars = 3395 km; radiusof the orbit of mars = 2.28 × 108kg; G= 6.67 × 10–11 m2kg–2.
Answer - 24 : -
Mass of the spaceship, ms = 1000 kg
Mass of the Sun, M = 2 × 1030 kg
Mass of Mars, mm = 6.4 × 10 23 kg
Orbital radius of Mars, R =2.28 × 108 kg =2.28 × 1011m
Radius of Mars, r = 3395 km= 3.395 × 106 m
Universal gravitational constant, G = 6.67 ×10–11 m2kg–2
Potentialenergy of the spaceship due to the gravitational attraction of the Sun 
Potential energy of thespaceship due to the gravitational attraction of Mars
Since the spaceship isstationed on Mars, its velocity and hence, its kinetic energy will be zero.
Totalenergy of the spaceship 
The negative signindicates that the system is in bound state.
Energy required forlaunching the spaceship out of the solar system
= – (Total energy of thespaceship)
A rocket is fired ‘vertically’ from thesurface of mars with a speed of 2 km s–1. If 20% of its initial energy is lostdue to Martian atmospheric resistance, how far will the rocket go from thesurface of mars before returning to it? Mass of mars = 6.4× 1023 kg; radius ofmars = 3395 km; G = 6.67× 10-11 N m2 kg–2.
Answer - 25 : -
Initial velocity of the rocket, v =2 km/s = 2 × 103 m/s
Mass of Mars, M = 6.4 × 1023 kg
Radius of Mars, R = 3395 km= 3.395 × 106 m
Universal gravitational constant, G = 6.67×10–11 N m2 kg–2
Mass of the rocket = m
Initialkinetic energy of the rocket = 
Initial potential energy of the rocket
Total initial energy
If 20 % of initial kineticenergy is lost due to Martian atmospheric resistance, then only 80 % of itskinetic energy helps in reaching a height.
Totalinitial energy available 
Maximum height reached by the rocket = h
At this height, thevelocity and hence, the kinetic energy of the rocket will become zero.
Totalenergy of the rocket at height h 
Applying the law ofconservation of energy for the rocket, we can write:
