An earth’s satellite near the surface of the earth takes about 90 min per revolution. A satellite orbiting the moon also takes about 90 min per revolution. Then which of the following is true ? [where ρm is density of the moon and ρe is density of the earth.]
ρm = ρe
For a satellite skimming the surface of a spherical body, the gravitational pull of the body supplies the centripetal force, and the orbit radius is practically the radius R of the body itself.
Step 1 — Equating gravity with the centripetal requirement: \(\dfrac{GM}{R^{2}}=\dfrac{4\pi^{2}R}{T^{2}}\).
Step 2 — Writing the mass through the mean density, \(M=\dfrac{4}{3}\pi R^{3}\rho\), and substituting gives \(T^{2}=\dfrac{4\pi^{2}R^{3}}{GM}=\dfrac{3\pi}{G\rho}\).
Step 3 — So \(T=\sqrt{\dfrac{3\pi}{G\rho}}\): the period of a near-surface orbit depends only on the density of the central body, not on its size or mass separately.
Step 4 — Both satellites take about 90 min, so the two densities must be equal, \(\rho_{m}=\rho_{e}\).
A smaller or larger lunar density would make the lunar period longer or shorter than the earth value, which contradicts the equal periods. A definite conclusion is possible because the period fixes the density uniquely.
ρm = ρe option is correct.
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