Satellite OrbitsJEE Main

Orbital & escape speed, Kepler's third law — interactive Physics simulation for IIT-JEE.

Concept

A satellite stays in orbit because gravity supplies exactly the centripetal force it needs. Balancing the two gives a unique orbital speed at each radius. Reaching escape speed ( $\sqrt2$ times larger) lets it leave the body entirely.

Key formula

v_o = \sqrt{\frac{GM}{r}}, \quad v_e = \sqrt{\frac{2GM}{r}} = \sqrt2\,v_o, \quad T = 2\pi\sqrt{\frac{r^3}{GM}}

Derivation

Set gravity equal to the centripetal requirement: $\dfrac{GMm}{r^2} = \dfrac{mv_o^2}{r}$ , giving $v_o = \sqrt{GM/r}$ .

Escape speed comes from energy: $\tfrac12 mv_e^2 = \dfrac{GMm}{r}$ , so $v_e = \sqrt{2GM/r}$ .

The period follows from $v_o = 2\pi r/T$ , yielding Kepler's third law $T^2 \propto r^3$ .

Scenarios to explore

Satellite Orbits — Orbital & escape speed, Kepler's third law.

Real-world applications

Geostationary communication satellites (T = 1 sidereal day).
Low-Earth-orbit imaging satellites (period ≈ 90 min).
Planning interplanetary launch windows.

JEE exam tips

$v_o$ at the surface ( $h=0$ ) for Earth ≈ 7.9 km/s; $v_e$ ≈ 11.2 km/s.
Total orbital energy $E = -\dfrac{GMm}{2r}$ — negative means bound.

Common mistakes

Using altitude instead of orbital radius $r = R + h$ from the centre.
Forgetting orbital speed decreases with altitude (higher = slower).
Mixing up $v_e = \sqrt2\,v_o$ direction — escape is the larger one.

Exam traps to avoid

Orbital speed is independent of the satellite's mass.
A higher orbit has more total energy (less negative) yet a smaller speed.