MHT CET · Physics · Gravitation
A body starts from rest from a distance \(\mathrm{R}_0\) from the centre of the earth. The velocity acquired by the body when it reaches the surface of the earth will be \((\mathrm{R}=\) radius of earth, \(\mathrm{M}=\) mass of earth \()\)
- A \(2 \mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)\)
- B \(\sqrt{2 \mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)}\)
- C \(\mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)\)
- D \(2 \mathrm{GM} \sqrt{\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)}\)
Answer & Solution
Correct Answer
(B) \(\sqrt{2 \mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)}\)
Step-by-step Solution
Detailed explanation
According to law of conservation of energy,
\(\frac{1}{2} m v^2=-\frac{\mathrm{GMm}}{\mathrm{R}_0}-\left(\frac{\mathrm{GMm}}{\mathrm{R}}\right)=\mathrm{GMm}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)\)
\(\therefore \quad \mathrm{v}^2=2 \mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)\)
\(\therefore \quad\) The velocity acquired by the body when it reaches the surface of earth is:
\(\mathrm{v}=\sqrt{2 \mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)}\)
Amongst the given options, only option (B) dimensionally equates to velocity.
\(\frac{1}{2} m v^2=-\frac{\mathrm{GMm}}{\mathrm{R}_0}-\left(\frac{\mathrm{GMm}}{\mathrm{R}}\right)=\mathrm{GMm}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)\)
\(\therefore \quad \mathrm{v}^2=2 \mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)\)
\(\therefore \quad\) The velocity acquired by the body when it reaches the surface of earth is:
\(\mathrm{v}=\sqrt{2 \mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)}\)
Amongst the given options, only option (B) dimensionally equates to velocity.
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