Question

A satellite is orbiting just above the surface of the earth with period T. If d is the dependent of the earth and G is the universal constant of gravitation, the quantity 3π/Gd respectively:

• A. T²
• B. T³
• C. \(\sqrt{T}\)
• D. T

Answer 1

The Correct Option is A

To solve this physics problem, we need to determine which quantity among the given options is equivalent to \(3\pi/Gd\) based on the orbital period of a satellite, denoted as \(T\), orbiting just above the Earth's surface. Let's break down the problem step-by-step:

1. First, let's recall Kepler's third law for a satellite orbiting a planet:
   • The formula is given by \(T^2 = \frac{4\pi^2 R^3}{GM}\), where \(R\) is the radius of the orbit (which is approximately the radius of the Earth for a satellite orbiting just above the surface), \(G\) is the gravitational constant, and \(M\) is the mass of the Earth.
2. Since the satellite is orbiting just above the Earth's surface, we can use the radius of the Earth, \(R = d\), where \(d\) is the radius of the Earth. Plug this into the above equation:
   • \(T^2 = \frac{4\pi^2 d^3}{GM}\)
3. From this expression, we can isolate the quantity and rearrange it to match the form in question:
   • We aim to find which option corresponds to \(3\pi/Gd\). From Kepler's law, we find \(\frac{3\pi}{Gd} \approx \frac{T^2}{R}\) if we consider dimensional analysis.
   • Thus, we equate \(R \approx d\) (radius of the Earth).
4. Therefore, we find that:
   • The correct answer option is T² since the formula we derived involves square of the period \(T\).

This demonstrates that the expression \(3\pi/Gd\) is proportional to \(T^2\) in the context of a satellite orbit above the Earth's surface using Kepler's third law and the dimensions involved.