Orbital Velocity Numericals
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Direct answer
Just above the Earth's surface a satellite must move at about 7.9 km/s to stay in orbit — the orbital velocity v0 = √(GM/r) = √(gR²/(R + h)), obtained by setting gravity equal to the centripetal requirement, GMm/r² = mv0²/r. The companions follow at once: escape velocity ve = √(2GM/R) = √2 × v0 = 11.2 km/s from the surface; time period T = 2π√(r³/GM), Kepler's third law in numbers; and the energy trio — potential −GMm/r, kinetic +GMm/2r, total −GMm/2r. A grazing orbit takes about 85 minutes; r ≈ 42200 km (height about 36000 km) gives T = 24 h, the geostationary condition used by India's communication satellites.
What you must remember
- Orbital velocity: v0 = √(GM/r); near the surface v0 = √(gR) ≈ 7.9 km/s with g = 9.8 m/s² and R = 6.4 × 10^6 m; it decreases as height increases.
- Escape velocity: ve = √(2gR) = 11.2 km/s for Earth; at any point, ve = √2 × v0 — multiplying orbital speed by 2 instead of √2 is the standard error.
- Time period: T = 2π√(r³/GM), so T² ∝ r³; close orbit about 85 min, geostationary 24 h at r ≈ 42200 km (h ≈ 36000 km) above the equator.
- Energy trio: PE = −GMm/r, KE = +GMm/2r, E = −GMm/2r; KE = |E| = ½|PE| — a favourite ratio.
- Anchor constants: G = 6.67 × 10^-11, M_Earth = 6 × 10^24 kg, R_Earth = 6.4 × 10^6 m, g = 9.8 m/s²; GM = gR² = 4 × 10^14 approx — the computational shortcut.
- Weightlessness decoded: the astronaut and satellite fall together toward Earth (apparent weight zero), though gravity at 36000 km is still about 0.22 m/s² — gravity is small, never absent.
- Binding logic: E negative means bound; raising the orbit adds energy (E becomes less negative); E = 0 marks escape.
One satellite, every relation
Place a 1000 kg satellite at height h = R above the surface, so r = 2R = 1.28 × 10^7 m. Orbital speed: v0 = √(GM/2R) = surface value/√2 ≈ 5.6 km/s. Period by scaling: T grows as r^(3/2), so 85 min × 2^1.5 ≈ 240 min — four hours. Energy: E = −GMm/2r = −(4 × 10^14 × 1000)/(2 × 1.28 × 10^7) ≈ −1.6 × 10^10 J. Each step used a scaling relation rather than a fresh computation, and that is exactly the pace NEET demands. The same satellite lifted to geostationary height would slow further, lengthen its day to 24 hours, and its energy would rise toward zero — less negative as r grows, which reads as "energy must be supplied", not lost.
Where NEET sets the trap
The deadliest slip is r versus h: formulas take r, the distance from Earth's centre, while questions hand you the height above the surface — every option list includes the answer computed with h instead of r. The √2 relation between escape and orbital speed is tested both numerically and as an assertion ("doubling the orbital speed gives escape" — false; 1.414 times is enough). Energy signs are mined: total energy is negative, kinetic positive, and "energy of a satellite" without a qualifier usually means the total. Geostationary questions bundle three conditions — 24-hour period, equatorial plane, and orbiting west to east with Earth's rotation — and each alone has been the credited answer to some question. Finally, "what happens if the satellite's speed suddenly drops to zero" (it falls, along a radius) separates vector thinkers from formula pluggers.
Frequently asked questions
What is the orbital velocity just above the Earth's surface?
v0 = √(gR) ≈ 7.9 km/s using g = 9.8 m/s² and R = 6.4 × 10^6 m; real low orbits run slightly slower because height is not zero.
How is escape velocity related to orbital velocity?
ve = √2 × v0 at the same distance — 11.2 km/s versus 7.9 km/s at the surface; the factor is √2, not 2.
At what height does a satellite become geostationary?
About 36000 km above the equator (r ≈ 42200 km from the centre), where T = 2π√(r³/GM) works out to 24 hours, matching Earth's rotation.
Why does an astronaut in orbit feel weightless?
Satellite and astronaut are both in free fall with the same acceleration, so neither presses on the other; gravity at that altitude is reduced but never zero.
What is the total energy of a satellite of mass m in an orbit of radius r?
E = −GMm/2r, the sum of KE (+GMm/2r) and PE (−GMm/r); the negative value marks a bound orbit, and supplying +GMm/2r would free it.