A satellite has a mass of 6182 kg and is in a circular orbit 4.43 × 105 m above the surface of a planet. The period of the orbit is 2.1 hours. The radius of the planet is 4.54 × 106 m. What would be the true weight of the satellite if it were at rest on the planet’s surface?
+33653 centripetal force provided by gravitational interaction:
\(\frac{mv^2}{r+h}=\frac{GMm}{(r+h)^2}\)
m = satellite mass, M = planet mass, r = planet radius, h = satellite height, G = gravitational constant, v = satellite velocity.
Rearrange: \(GM=v^2(r+h)\)
Satellite velocity: \(v=\frac{2*\pi(r+h)}{T}\) T = orbital period.
Hence: \(GM=\frac{(2\pi)^2(r+h)^3}{T^2}\)
Weight on surface: \(W=\frac{GMm}{r^2}\rightarrow \frac{(2\pi)^2m(r+h)^3}{r^2T^2}\)
With m in kilograms, r and h in metres, T in seconds, then W is in Newtons.
I'll leave you to do the number crunching.
