Gravitation Jee Advanced

Question

A small satellite is moving in a circular orbit around a massive
spherical planet of radius R. A small asteroid of mass same as that of
the satellite starts from rest from a large distance from the planet in a
radial manner and in its journey it collides with the satellite and gets
stuck to it. The combined new body now revolves around the planet
in a new orbit which just grazes the planet at some point. Find the
radius of the original circular orbit of the satellite.
Assume:
Mass of planet=M
Mass of satellite=m

Answer 1

• We are free to conserve energy and angular momentum(As force is radial) also we are free to conserve momentum.
   
• Key concepts: Angular momentum conservation, Momentum Conservation, Energy Conservation, Orbital velocity, Elliptical orbit.
• We firstly conserve Momentum for the asteroid and satellite system;

          Orbital velocity=\(\sqrt{\frac{GM}{x}}=V\) (for satellite)

• By energy conservation for asteroid, we get velocity just before collision \(\sqrt{\frac{2GM}{x}}= V\sqrt{2}\)
• Now we conserve linear momentum for satellite asteroid system,
  Asteroid has velocity radially inwards and Satellite velocity tangential.
  so we get net velocity of satellite asteroid system \(V=(\sqrt{3}/2)\)
   
• The asteroid satellite system grazes earth so we can conserve angular momentum for the system (as it will graze in elliptical orbit)
  \((V/2)x=V_{1}R\) (\(V_{1} \) is the velocity where it grazes earth)----------(1)
• We now conserve energy to get our unknowns 
  \(\frac{1}{2}\frac{2m}{1}(\frac{v\sqrt{3}}{2})^2\)-\(\frac{2GMm}{x}\)=\(\frac{2m({V_1})^2}{2}\)-\(\frac{2GMm}{R}\)----------(2)
  
  We get the equation 
  \(x^2-8Rx+5R^2=0\)
  
  
  So the the original radius of the orbit was 
  X=\(R(4+\sqrt{11})\)

Comments

Explained very fruitfully