Gravitation: Real-World Scenarios (4)

Question

ISRO’s Chandrayaan-3 orbited the Moon at an altitude of about $100$ km in a circular orbit. Calculate the orbital speed of the spacecraft. (Take Moon’s mass $M = 7.35 \times 10^{22}$ kg, Moon’s radius $R = 1.74 \times 10^6$ m, $G = 6.67 \times 10^{-11}$ N·m²/kg².) This is a NEET 2023 pattern problem rephrased with real ISRO context.

Solution — Step by Step

Why This Works

A circular orbit is a balance: gravity pulls the satellite toward the Moon, and the satellite’s tangential motion keeps it from falling in. At the right speed, these match exactly, and the spacecraft traces a circle.

Notice the orbital speed depends only on the central body’s mass and the orbital radius — not on the spacecraft’s mass. Chandrayaan-3 and a 1-tonne dummy at the same altitude would orbit at the same speed.

Alternative Method

Use $g_{\text{moon}}$ at altitude. The acceleration due to gravity at radius $r$ is $g(r) = GM/r^2$. For circular orbit, $v = \sqrt{g(r) \cdot r}$. Compute $g(r) \approx 1.45$ m/s², so $v = \sqrt{1.45 \times 1.84 \times 10^6} \approx 1631$ m/s. Same answer.