The minimum and maximum distances of a satellite from the center of the Earth are 2R and 4R respectively, where R is the radius of Earth and M is the mass of Earth. The radius of curvature at the point of minimum distance is

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Solution

Step 1: Formula used
By conservation of angular momentum,
$m v_{1} R_{1} = m v_{2} R_{2} \Rightarrow v_{1} R_{1} = v_{2} R_{2} \Rightarrow v_{1} (2 R) = v_{2} (4 R) \Rightarrow v_{1} = 2 v_{2} \Rightarrow v_{2} = \frac{v_{1}}{2} \to (1)$
$R_{1} \to$ Minimum distance of satellite from the center of earth
$R_{2} \to$ Maximum distance of satellite from the center of earth
$m \to$ Mass of the satellite
$v_{1} \to$ Velocity of satellite at minimum distance
$v_{2} \to$ Velocity of satellite at maximum distance
Step 2: Use conservation of energy and substitute the values
By conservation of energy
Total energy = Kinetic energy + Potential energy
${(K i n e t i c e n e r g y + P o t e n t i a l e n e r g y)}_{a t m i n d i s \tan c e} = {(K i n e t i c e n e r g y + P o t e n t i a l e n e r g y)}_{a t m a x d i s \tan c e}$
$- \frac{G M m}{2 R} + \frac{1}{2} m v_{1}^{2} = - \frac{G M m}{4 R} + \frac{1}{2} m v_{2}^{2} \to (2)$
Substitute $(1)$ in $(2)$ , we get
$v_{1} = \sqrt{\frac{G M}{6 R}} v_{2} = \sqrt{\frac{2 G M}{3 R}}$
Step 3: Find the radius of curvature
If r is the radius of curvature at the minimum distance,
$\frac{m v_{1}^{2}}{r} = \frac{G M m}{{(2 R)}^{2}} \Rightarrow r = \frac{4 v_{1}^{2} R^{2}}{G M} \Rightarrow r = \frac{8 R}{3} (S u b s t i t u t i n g v a l u e o f v_{1})$
The radius of curvature at the point of minimum distance is $r = \frac{8 R}{3}$ .

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