A body starts from rest from a distance $\mathrm{R}_0$ from the centre of the earth. The velocity acquired…

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 $)$
  1. $\quad 2 \mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)$
  2. $\sqrt{2 \mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)}$
  3. $\mathrm{GM}\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)$
  4. $\quad 2 \mathrm{GM} \sqrt{\left(\frac{1}{\mathrm{R}}-\frac{1}{\mathrm{R}_0}\right)}$

Solution

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.

Asked in: MHT CET 2024 (10 May Shift 2)

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