Question

Consider two satellites A and B of equal mass m, moving in the same circular orbit about the earth, but in the opposite sense as shown in figure. The orbital radius is r. The satellites undergo a collision which is perfectly inelastic. For this situation, mark out the correct statement(s). (Take mass of earth as M)

• A. The total energy of the two satellites plus earth system just before collision is $-\left(GMm\right)/r$.
• B. The total energy of the two satellites plus earth system just after collision is $-\left(2GMm\right)/r$.
• C. The total energy of two satellites plus earth system just after collision is $-\left(GMm\right)/2r.$
• D. The combined mass (two satellites) will fall towards the earth just after collision.

Answer 1

Answer: (A), (B), (D)

The total energy of the two satellites plus earth system just before collision is $-\left(GMm\right)/r$.
The total energy of the two satellites plus earth system just after collision is $-\left(2GMm\right)/r$.
The combined mass (two satellites) will fall towards the earth just after collision.

Just before collision, the total energy of one satellite is KE+PE i.e
$E=\frac{GMm}{2r}-\frac{GMm}{r}=-\frac{GMm}{2r}$
Thus for two satellites it will be $-\frac{GMm}{r}$
Let orbital velocity be v, then from momentum conservation,
$mv-mv=2m\times v_1$
$v_1=0$
As velocity of combined mass just after collision is zero it has no kinetic energy. At this instant, the total energy of the system only consists of the gravitational potential energy given by
$U=-\frac{GM\times 2m}{r}$