Q62 / 75 · JEE Main 2025 (28 Jan Shift 2) · Physics

Earth has mass 8 times and radius 2 times that of a planet. If the escape velocity from the earth is $11.2 \mathrm{~km} / \mathrm{s}$, the escape velocity in $\mathrm{km} / \mathrm{s}$ from the planet will be :

  • A
    $2.8$
  • B
    $11.2$
  • Correct answer
    C
    $5.6$
  • D
    $8.4$

Solution

$\begin{aligned} & \mathrm{V}_{\text {escape }}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}} \\ & \frac{\left(\mathrm{V}_{\text {escape }}\right)_{\text {Planet }}}{\left(\mathrm{V}_{\text {escape }}\right)_{\text {Earth }}}=\sqrt{\left(\frac{\mathrm{M}_{\mathrm{P}}}{\mathrm{M}_{\mathrm{E}}}\right) \times\left(\frac{\mathrm{R}_{\mathrm{E}}}{\mathrm{R}_{\mathrm{P}}}\right)}=\frac{1}{2} \\ & \left(\mathrm{~V}_{\text {escape }}\right)_{\text {Planet }}=\frac{1}{2}\left(\mathrm{~V}_{\text {escape }}\right)_{\text {Earth }}=5.6 \mathrm{~km} / \mathrm{s}\end{aligned}$

Hence, the correct answer is option: 'C'.

Relevant Questions