MHT CET · Physics · Gravitation
A body is taken to a height of \(n R\) from the surface of the earth. The ratio of the acceleration due to gravity on the surface to that at the altitude is
- A \((n+1)^{2}\)
- B \((n+1)^{-2}\)
- C \((n+1)^{-1}\)
- D \((n+1)\)
Answer & Solution
Correct Answer
(A) \((n+1)^{2}\)
Step-by-step Solution
Detailed explanation
Aceleration due to gravity at a height \(h\) above
the earth surface \(g^{\prime}=g\left(\frac{R}{R+h}\right)^{2}\)
\(
\begin{array}{l}
\frac{g}{g^{\prime}}=\left(\frac{R+h}{R}\right)^{2} \\
\frac{g}{g^{\prime}}=\left(\frac{R+n R}{R}\right)^{2} \\
\frac{g}{g^{\prime}}=(1+n)^{2}
\end{array}
\)
the earth surface \(g^{\prime}=g\left(\frac{R}{R+h}\right)^{2}\)
\(
\begin{array}{l}
\frac{g}{g^{\prime}}=\left(\frac{R+h}{R}\right)^{2} \\
\frac{g}{g^{\prime}}=\left(\frac{R+n R}{R}\right)^{2} \\
\frac{g}{g^{\prime}}=(1+n)^{2}
\end{array}
\)
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