Planet and Satellite: Practice Problem & Solution
For energy of satellite, match the columns (symbols have their respective meaning): **Column-I** (i) Kinetic energy (ii) Potential energy (iii) Total energy **Column-II** (p) \(\frac{L^2}{2mr^2}\) (q) \(-\frac{L^2}{mr^2}\) (r) \(-\frac{L^2}{2mr^2}\)
Solution Explained:
To solve this problem, we apply the core principles of Planet and Satellite. Understanding the underlying formula is key to arriving at the correct answer below:
Kinetic energy \(K = \frac{1}{2}mv^2 = \frac{L^2}{2mr^2}\) (since \(L = mvr\)). Potential energy \(U = -\frac{GMm}{r} = -\frac{L^2}{mr^2}\). Total energy \(E = K + U = -\frac{L^2}{2mr^2}\). Thus (i)-p, (ii)-q, (iii)-r.
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