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 determine how the quantity \( \frac{3\pi}{Gd} \) relates to the satellite's orbital period \( T \), we begin with relevant physics principles. The satellite is orbiting just above the Earth's surface, implying it is in a low Earth orbit. The gravitational force is the centripetal force required to keep the satellite in orbit. Using the formula for gravitational force \( F = \frac{G \cdot M \cdot m}{r^2} \) and centripetal force \( F = \frac{m \cdot v^2}{r} \), equating both gives:

\(\frac{G \cdot M \cdot m}{r^2} = \frac{m \cdot v^2}{r}\)

Cancel \( m \) and \( r \) from both sides:

\(v^2 = \frac{G \cdot M}{r}\)

The relationship between velocity \( v \) and orbital period \( T \) is \( v = \frac{2\pi r}{T} \), substituting gives:

\(\left(\frac{2\pi r}{T}\right)^2 = \frac{G \cdot M}{r}\)

Simplify:

\(\frac{4\pi^2 r^2}{T^2} = \frac{G \cdot M}{r}\)

Rearranging terms leads to:

\(T^2 = \frac{4\pi^2 r^3}{G \cdot M}\)

For a satellite just above Earth’s surface, \( r \) is approximately equal to Earth's radius \( R \), thus:

\(T^2 \propto \frac{r^3}{G \cdot M}\)

Given \( \frac{3\pi}{Gd} \) and comparing it with the derived \( T^2 \), it suggests that \( \frac{3\pi}{Gd} \) is proportional to \( T^2 \), confirming the relationship. Therefore, the correct answer is \( T^2 \).