For any satellite of earth, its period [math: T] is related to its radius of orbit [math: R] by the equation [math: R3=KT2] R^{3}=KT^{2} , where [math: K] is a constant. If [math: G] is the gravitational constant and [math: M] is the mass of the earth, which one of the following is equal to [math: K] for any earth satellite?

Question

BlackTom AI · Accepted Answer

We start by recognizing the standard relation for a satellite in orbit: T^2 is proportional to R^3, with the proportionality constant involving G and M. The problem states R^3 = K T^2, so K must be the factor that converts T^2 to R^3.

First, recall the classical formula for orbital motion in a gravitational field: T^2 = (4π^2/GM) R^3. This expresses the well-known dependence of the orbital period on radius for a small body orbiting a mass M with gravitational constant G.

If we rewrite this equation to isolate R^3, we get R^3 = (GM/4π^2) T^2. Here, the constant multiplying T^2 is exactly GM/(4π^2).

Now, evaluate each option:
- Option a: 2/√(GM). This expression has units that do not match the required proportionality between R^3 and T^2 and does not involve T^2 at all in a proper multiplicative way. It also lacks the T^2 factor entirely, so it cannot be the correct K.
- Option b: GM/(4π^2). This matches the derived form R^3 = (GM/4π^2) T^2, with the correct dependence on GM and the 4π^2 factor in the denominator. It is dimensionally consistent and aligns with the standard gravity-based derivation.
- Option c: 2πGM. This is missing the necessary division by 4π^2 and introduces an extra factor of 2π, which would disrupt the required T^2 to R^3 relationship; it cannot be correct.
- Option d: 4π^2GM. This places GM multiplied by 4π^2 in the numerator, which again does not match the derived coefficient GM/(4π^2); the factor is inverted, so it is incorrect.

Putting it together, the only option that reproduces the established relation R^3 = (GM/4π^2) T^2 is the one that states K = GM/(4π^2).

MUF0121 Physics Unit 1 - Semester 2, 2025

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