A satellite of mass m=500 kg orbits the Earth at a height h=600 km above the surface in a circular orbit. Take RE=6400 km, g=9.8 m/s². Find (a) its orbital speed, (b) the time period of revolution and (c) the total mechanical energy of the satellite.
Solution — Step by Step
Orbital radius r=RE+h=6400+600=7000 km =7×106 m. Acceleration due to gravity at orbit:
g′=g(rRE)2=9.8×(70006400)2≈9.8×0.836≈8.19 m/s2
For circular motion, rmv2=mg′, so
v=g′r=8.19×7×106≈7.57×103 m/s≈7.57 km/sT=v2πr=7.57×1032π×7×106≈5810 s≈96.8 min
For a circular orbit, E=−2rGMm=−21mv2 (the kinetic and potential energies have a fixed ratio K=−E, U=2E).
E=−21(500)(7570)2≈−1.43×1010 J
Orbital speed ≈7.57 km/s, period ≈96.8 min, energy ≈−1.43×1010 J.
Why This Works
Two ideas drive every satellite numerical: gravity provides the centripetal force, and total energy in a bound orbit is negative. The negative sign is not a calculation error — it tells us the satellite is bound to Earth and would need positive energy added to escape.
The trick of using g′=g(R/r)2 avoids ever computing GME directly. Examiners love this because it tests whether you remember that g at the surface already encodes GME/R2.
Alternative Method
You can compute GME=gRE2=9.8×(6.4×106)2≈4.01×1014 m³/s². Then v=GME/r directly. Same result, slightly more typing.
Common Mistake
Students plug h=600 km directly into r instead of r=RE+h. This single error has been the most common mistake on NEET 2022 satellite problems — the answer comes out roughly 4× too fast and gets you zero. Always draw a quick diagram with the Earth’s centre marked.
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