Gravitation: Conceptual Doubts Cleared (6)

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Question

A satellite is in a circular orbit at altitude h=Rh = R above Earth’s surface (where RR is Earth’s radius). Why doesn’t it fall, even though gravity is the only force acting on it? And what would happen if its speed suddenly doubled?

Solution — Step by Step

The satellite is falling — every second. It’s just falling sideways fast enough that the curved path matches Earth’s curvature. Gravity provides the centripetal force; it doesn’t oppose motion, it bends the trajectory.

For circular orbit at radius r=R+h=2Rr = R + h = 2R:

GMmr2=mvo2r    vo=GMr=GM2R\frac{GMm}{r^2} = \frac{mv_o^2}{r} \implies v_o = \sqrt{\frac{GM}{r}} = \sqrt{\frac{GM}{2R}}

Escape velocity at radius rr is ve=2GMr=2vov_e = \sqrt{\tfrac{2GM}{r}} = \sqrt{2}\, v_o. So if orbital speed doubles, the new speed is 2vo>2vo2v_o > \sqrt{2}\, v_o.

The satellite has total energy E=12m(2vo)2GMmr=2mvo2mvo2=+mvo2>0E = \tfrac{1}{2}m(2v_o)^2 - \tfrac{GMm}{r} = 2mv_o^2 - mv_o^2 = +mv_o^2 > 0. Positive total energy means the orbit becomes a hyperbola — the satellite escapes Earth permanently.

Final answer: It doesn’t fall because it’s in free-fall along a curved path. If speed doubles, it escapes on a hyperbolic trajectory.

Why This Works

Orbital motion is the cleanest example of “circular motion under a single radial force”. The mistake is to imagine gravity as a force pulling the satellite “down” — there is no down in space. Gravity always points toward Earth’s centre, and the satellite’s tangential velocity is perpendicular to it.

The total energy criterion (E<0E < 0 bound, E=0E = 0 parabolic escape, E>0E > 0 hyperbolic escape) is the gold standard for predicting orbital fate.

Alternative Method

Compare kinetic energy directly to the binding energy Ebound=GMm2r|E_{\text{bound}}| = \tfrac{GMm}{2r}. If KE>KE > binding energy, the satellite escapes. Doubling vv quadruples KEKE, easily exceeding the binding threshold.

Remember: ve=2vov_e = \sqrt{2}\, v_o at any altitude. JEE Main 2023 had this exact ratio question — many students wasted time deriving it from scratch.

Common Mistake

Students assume “doubling speed means double the radius”. Wrong — orbital radius depends on v2v^2, but the new orbit isn’t even closed. Always check total energy sign before declaring a new circular orbit.

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