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KCET · Physics · Gravitation

The acceleration due. to gravity becomes \(\left(\frac{g}{2}\right)\) \((g=\) acceleration due to gravity on the surface of the earth) at a height equal to

  1. A \(4 R\)
  2. B \(\frac{R}{4}\)
  3. C \(2 R\)
  4. D \(\frac{R}{2}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\frac{R}{4}\)

Step-by-step Solution

Detailed explanation

The acceleration due to gravity
\(g=\frac{G M}{R^{2}}\)
At a height \(\mathrm{h}\) above the earth's surface, the acceleration due to gravity is
\(g^{\prime} =\frac{G M}{(R+h)^{2}} \)
\(\frac{g}{g^{\prime}} =\left(\frac{R+h}{R}\right)^{2} \)
\(=\left(1+\frac{h}{R}\right)^{2} \)
\(\frac{g^{\prime}}{g} =\left(1+\frac{h}{R}\right)^{-2} \)
\(=\left(1-\frac{2 h}{R}\right) \)
\(\text {but } \quad \frac{g / 2}{h} =\frac{g}{2} (given) \)
\(\therefore \frac{2 h}{R} =\frac{1}{2} \)
\(h =\frac{R}{4}\)
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