The escape velocity for a planet whose radius is \(1.7 \times 10^6 \mathrm{~m}\) and acceleration due to…

The escape velocity for a planet whose radius is \(1.7 \times 10^6 \mathrm{~m}\) and acceleration due to gravity is \(1.7 \mathrm{~ms}^{-2}\) is
  1. \(1.7 \mathrm{kms}^{-1}\)
  2. \(2.89 \mathrm{kms}^{-1}\)
  3. \(1.7 \sqrt{2} \mathrm{kms}^{-1}\)
  4. \(3.4 \mathrm{kms}^{-1}\)

Solution

Radius of planet, \(R=1.7 \times 10^6 \mathrm{~m}\) Acceleration due to gravity, \(g=1.7 \mathrm{~ms}^{-2}\) \(\therefore\) Escape velocity on the surface of planet is given as \(\begin{aligned} v_e & =\sqrt{2 g R}=\sqrt{2 \times 1.7 \times 1.7 \times 10^6} \\ & =1.7 \sqrt{2} \times 10^3 \mathrm{~ms}^{-1}=1.7 \sqrt{2} \mathrm{~km} \mathrm{~s}^{-1} \end{aligned}\)

Asked in: AP EAMCET 2020 (18 Sep Shift 1)

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