AP EAMCET · PHYSICS · Gravitation
The escape speed of an object on the surface of the earth is \(\mathrm{V}\). If the object is thrown out with speed \(4 \mathrm{~V}\) from the surface of the earth, the speed of the object far away from the earth is
- A \(3 \mathrm{~V}\)
- B \(\sqrt{15} \mathrm{~V}\)
- C \(2.5 \mathrm{~V}\)
- D \(\sqrt{8} \mathrm{~V}\)
Answer & Solution
Correct Answer
(B) \(\sqrt{15} \mathrm{~V}\)
Step-by-step Solution
Detailed explanation
Let \(V_0\) be the speed of object far away from earth. By law of conservation of mechanical energy \(\frac{1}{2} \mathrm{~m}(\mathrm{UV})^2-\frac{\mathrm{GMm}}{\mathrm{R}}=\frac{1}{2} \mathrm{mV}_0^2+0\)…
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