Question

Two satellites P and Q are moving in different circular orbits around the Earth (radius 𝑅). The heights of P and Q from the Earth surface are \(ℎ_P\) and \(ℎ_Q\), respectively, where \(ℎ_P = \frac{𝑅}{3}\). The accelerations of P and Q due to Earth’s gravity are \(𝑔_P\) and \(𝑔_Q\), respectively. If \(\frac{𝑔_P}{𝑔_Q} = \frac{36}{25}\), what is the value of \(ℎ_Q?\)

• A. \(\frac{3𝑅}{5}\)
• B. \(\frac{𝑅}{6}\)
• C. \(\frac{6𝑅}{5}\)
• D. \(\frac{5𝑅}{6}\)

Answer 1

The Correct Option is A

To solve this problem, we need to determine the height \(h_Q\) of satellite Q above the Earth's surface. Given:

\(h_P = \frac{R}{3}\) 

\(\frac{g_P}{g_Q} = \frac{36}{25}\)

We know that the gravitational acceleration \(g\) at a height \(h\) is given by:

\(g = \frac{g_0}{(1+\frac{h}{R})^2}\)

where \(g_0\) is the gravitational acceleration at the Earth's surface. For satellite P:

\(g_P = \frac{g_0}{(1+\frac{h_P}{R})^2}\)

\(g_P = \frac{g_0}{(1+\frac{R}{3R})^2} = \frac{g_0}{(\frac{4}{3})^2} = \frac{9g_0}{16}\)

For satellite Q:

\(g_Q = \frac{g_0}{(1+\frac{h_Q}{R})^2}\)

Using the relationship \(\frac{g_P}{g_Q} = \frac{36}{25}\), we get:

\(\frac{\frac{9g_0}{16}}{\frac{g_0}{(1+\frac{h_Q}{R})^2}} = \frac{36}{25}\)

This can be simplified to:

\(\frac{9}{16} \times (1+\frac{h_Q}{R})^2 = \frac{36}{25}\)

On further simplification,

\((1+\frac{h_Q}{R})^2 = \frac{16}{9} \times \frac{36}{25} = \frac{64}{25}\)

Taking the square root:

\(1+\frac{h_Q}{R} = \frac{8}{5}\)

\(\frac{h_Q}{R} = \frac{8}{5} - 1 = \frac{3}{5}\)

Hence:

\(h_Q = \frac{3R}{5}\)

The correct answer is therefore:

\(\frac{3R}{5}\)

Answer 2

Given \(h_p=\frac{R}{3}\)
\(h_q=?\)
gravitational acceleration at height
\(g_{ht}=\frac{GM}{(R+h)^2}\)

\(\frac{g_P}{g_Q}=\frac{36}{25}\)

\(=\frac{\frac{GM}{(R+h_p)^2}}{\frac{GM}{(R+h_Q)^2}}\)

when  \(h_P=\frac{R}{3}\)
 on solving we get 
\(h_Q=\frac{3R}{5}\)