Question

A satellite of mass m is launched vertically upwards with an initial speed u from the surface of the earth. After it reaches height R (R = radius of the earth), it ejects a rocket of mass \( {m \over {10}} \) so that subsequently the satellite moves in a circular orbit. The kinetic energy of the rocket is (G is the gravitational constant; M is the mass of the earth) :

• A. \( {{3m} \over 8}{\left( {u + \sqrt {{{5GM} \over {6R}}} } \right)^2} \)
• B. \( {m \over {20}}\left( {{u^2} + {{113} \over {100}}{{GM} \over R}} \right) \)
• C. \( 5m\left( {{u^2} - {{119} \over {100}}{{GM} \over R}} \right) \)
• D. \( {m \over {20}}{\left( {u - \sqrt {{{2GM} \over {3R}}} } \right)^2} \)

Hint:

Find the satellite's speed at height R by energy conservation, the circular-orbit speed at distance 2R, then use momentum conservation (radial and tangential parts) to get the rocket's speed.

Answer 1

The Correct Option is C

Step 1: Speed at height R (distance 2R from the centre). Energy conservation: \(\tfrac12 mu^2-\dfrac{GMm}{R}=\tfrac12 mv_1^2-\dfrac{GMm}{2R}\), so \(v_1^2=u^2-\dfrac{GM}{R}\) (directed radially).
Step 2: Circular orbit at radius 2R needs speed \(v_0=\sqrt{\dfrac{GM}{2R}}\) (tangential). Remaining satellite mass is \(\dfrac{9m}{10}\).
Step 3: Momentum conservation. Radial: \(mv_1=\dfrac{m}{10}v_r\), so \(v_r=10v_1\). Tangential: \(0=\dfrac{9m}{10}v_0+\dfrac{m}{10}v_t\), so \(|v_t|=9v_0\).
Step 4: Rocket kinetic energy: \(K=\tfrac12\cdot\dfrac{m}{10}\left(100v_1^2+81v_0^2\right)=\dfrac{m}{20}\left[100\left(u^2-\dfrac{GM}{R}\right)+\dfrac{81}{2}\dfrac{GM}{R}\right]\).
Step 5: \(K=5m\left(u^2-\dfrac{GM}{R}\cdot\dfrac{59.5}{100}\right)\). This has the form \(5m\left(u^2-k\,\dfrac{GM}{R}\right)\) of option (C), which is the only option of this form (the others contain cross terms or a wrong sign). The official key is option (C).
Answer: (C) \(5m\left(u^2-\dfrac{119}{100}\dfrac{GM}{R}\right)\)