Question

A satellite revolving round the earth at a height h from earth's surface. The satellite explodes into equal parts, one part has zero velocity after the explosion, and the other part moves in the same direction with double the orbital speed of the original satellite. The maximum height above the Earth's surface reached by the part with double velocity is (Neglect the gravitational effects between the fragments) :-
A. 3h + 4R
B. 3R + 4h
C. 3(R + h)
D. None of the above

Answer 1

The satellite executing circular motion, breaks into equal parts implies that the mass of the two parts is the same (say, m/2). The laws of conservation of angular momentum and conservation of energy can be applied to deal with these problems.

Formula used:
The total energy of a satellite exhibiting circular motion:
$E = \dfrac{1}{2}mv^2 - \dfrac{GMm}{r}$;
where the velocity of a satellite at a distance r from the centre of Earth (mass M):
$v = \sqrt{ \dfrac{GM}{r}}$.

Complete step by step answer:
Assume the initial mass of the satellite, revolving at a height h, to be m. Then the velocity for it can be written as:
$v = \sqrt{ \dfrac{GM}{R+h}}$.
The total energy in this case is:
$E = \dfrac{1}{2}mv^2 - \dfrac{GMm}{r}$.
The satellite then breaks into two parts, one with zero velocity, another with velocity 2v. The final total energy becomes:
$E = \dfrac{m}{4} (2v)^2 - \dfrac{GMm}{2r’}$;
where we denote the separation of part 2 from the centre of the earth as r’.

Conservation of energy gives:
$\dfrac{1}{2}mv^2 - \dfrac{GMm}{r} = \dfrac{m}{4} (2v)^2 - \dfrac{GMm}{2r’}$
We can simplify this as:
$\dfrac{1}{2}v^2 = \dfrac{GM}{2r’} - \dfrac{GM}{r}$
making r and r’ substitutions;
$\dfrac{1}{2}v^2 = \dfrac{GM}{2R+2h’} - \dfrac{GM}{R+h}$
$\dfrac{1}{2}v^2 = \dfrac{GMm}{R+h} \left( \dfrac{R+h}{2R+2h’} -1 \right)$
At this point we simplify the expression, by making a substation for v and canceling it on both sides. The final expression we get will be:
$1= \left( \dfrac{R+h}{2R+2h’} -2 \right)$
We are asked to find the maximum height h’ of the satellite above the earth. From the above relation, we may write:
$(R+h) = 3(R+h’)$ ;
So, we get
$h’ = \dfrac{h – 2R}{3}$.

So, the correct answer is “Option D”.

Note:
The 1st part of the satellite has velocity zero. This conveys us the fact that it has zero kinetic energy and also it is no longer in the orbit of the earth. It must have burned or would have escaped the earth because from the formula of velocity, we can note that the velocity even at the surface of the earth is not zero.