AP EAMCET · PHYSICS · Gravitation
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
- A \(1.7 \mathrm{kms}^{-1}\)
- B \(2.89 \mathrm{kms}^{-1}\)
- C \(1.7 \sqrt{2} \mathrm{kms}^{-1}\)
- D \(3.4 \mathrm{kms}^{-1}\)
Answer & Solution
Correct Answer
(C) \(1.7 \sqrt{2} \mathrm{kms}^{-1}\)
Step-by-step Solution
Detailed explanation
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…
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