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TS EAMCET · Physics · Gravitation

A satellite revolving around the earth at a certain height experiences acceleration due to gravity equal to \(\frac{16}{49} g_0\), where \(g_0\) is the acceleration due to gravity on the earth's surface. If \(R\) is the radius of earth, then the square of time period of the satellite's revolution is equal to \(K\left[\frac{\pi^2 R^3}{G M}\right]\). The value of \(K\) is

  1. A \(\frac{27}{36}\)
  2. B \(\frac{343}{16}\)
  3. C \(\frac{125}{64}\)
  4. D \(\frac{675}{81}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{343}{16}\)

Step-by-step Solution

Detailed explanation

Acceleration due to gravity at height \(h\) from the surface of earth, \( \begin{aligned} g & =g_0\left(\frac{R}{R+h}\right)^2=\frac{16}{49} g_0 \\ \frac{R}{R+h} & =\frac{4}{7} \end{aligned} \) \(h=\frac{3 R}{4}\)....(i) Now, orbital velocity of satellite, \(v_0=\sqrt{G M / r}\)…
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