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MHT CET · Physics · Gravitation

A body (mass \(\mathrm{m}\) ) starts its motion from rest from a point distant \(R_0\left(R_0>R\right)\) from the centre of the earth. The velocity acquired by the body when it reaches the surface of earth will be ( \(\mathrm{G}=\) universal constant of gravitation, \(\mathrm{M}=\) mass of earth, \(\mathrm{R}\) = radius of earth)

  1. A \(2 \mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)\)
  2. B \(\left[2 \mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)\right]^{\frac{1}{2}}\)
  3. C \(\mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)\)
  4. D \(2 \mathrm{GM}\left[\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)\right]^{\frac{1}{2}}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(\left[2 \mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)\right]^{\frac{1}{2}}\)

Step-by-step Solution

Detailed explanation

According to law of conservation of energy,
\(\begin{aligned}
& \quad \frac{1}{2} \mathrm{mv}^2=-\frac{\mathrm{GMm}}{\mathrm{R}_0}-\left(\frac{\mathrm{GMm}}{\mathrm{R}}\right)=\mathrm{GMm}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right) \\
& \therefore \quad \mathrm{v}^2=2 \mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)
\end{aligned}\)
\(\therefore \quad\) The velocity acquired by the body when it reaches the surface of earth is:
\(\mathrm{V}=\left[2 \mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)\right]^{\frac{1}{2}}\)