Question

A satellite of mass m possessing a total energy U orbits round a planet of radius R at a height 'h'
from its surface in a circular path. It is suddenly stopped in its orbit and allowed to fall. The speed
with which it reaches the surface of planet is (h < R)​

Answer 1

Answer:

Total energy of the satellite T=

2r

−GMm

where r=R+h

∴ T=

2(R+h)

−GMm

Acceleration due to gravity at earth's surface g

o

=

R

2

GM

⟹GM=g

o

R

2

⟹ T=

2(R+h)

−mg

o

R

2

Answer 2

Answer:

The velocity at which satellite will fall on Earth, when suddenly stopped in its orbit is √(GM/R+h).

Explanation:

Given that :

• Total energy of satellite = U
• Satellite distance from the centre of Earth= ( R + h )

Solution :

• The total energy possessed by an orbiting satellite is

$Total \: energy = Potential \: Energy+ Kinetic \: Energy \\ = - \frac{GMm}{2r}$

• When the satellite stops orbiting the planet suddenly, the total energy changes into potential energy.
• When satellite is allowed to fall, this potential energy changes into kinetic energy.
• Let, the speed at which satellite reaches the Earth surface be v m/sec.

$\frac{GMm}{2(R+h)} = \frac{1}{2} m {v}^{2} \\ {v}^{2} = \frac{GM}{(R+h)} \\ v = \sqrt{ \frac{GM}{(R+h)} }$

• Hence, velocity at which satellite will fall on Earth is √(GM/R+h) .