An Earth satellite in a circular orbit of radius R has a period T. What is the period of an Earth satellite in a circular orbit of radius 4R ?

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BlackTom AI · Accepted Answer

We start from Kepler's third law for orbital motion around a central body like Earth: the orbital period squared is proportional to the semi-major axis cubed, i.e., T^2 ∝ r^3 for circular orbits. Let the initial orbit have radius R and period T, so T^2 ∝ R^3. For a second orbit with radius 4R, its period T' must satisfy T'^2 ∝ (4R)^3 = 64 R^3. Taking the ratio of the two equations: (T')^2 / T^2 = (64 R^3) / (R^3) = 64. Therefore T' = T * sqrt(64) = T * 8. In words, when the orbital radius increases by a factor of 4, the period increases by a factor of 8, due to the cubic dependence of radius in Kepler's law. This aligns with the general relation T ∝ r^(3/2) for circular orbits around a given central mass. If one were to check the dimensional consistency, time scales with the radius to the 3/2 power, so multiplying the radius by 4 multiplies the period by 4^(3/2) = 8. Consequently, the period for the 4R orbit is eight times the original period T.

An Earth satellite in a circular orbit of radius R has a period T. What is the period of an Earth satellite in a circular orbit of radius 4R ?

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