Question:
An object is thrown directly away from Earth's surface. Its initial speed is \(V_0\) and escape velocity is \(V_E\). If the object reaches a distance of \(\frac{4R}{3}\) from the centre of the Earth, then the ratio \(\frac{V_E}{V_0}\) is:
An object is thrown directly away from Earth's surface. Its initial speed is \(V_0\) and escape velocity is \(V_E\). If the object reaches a distance of \(\frac{4R}{3}\) from the centre of the Earth, then the ratio \(\frac{V_E}{V_0}\) is:
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For escape velocity type questions, always compare total mechanical energy at two radial positions using gravitational potential energy \(-\frac{GMm}{r}\).
Updated On: Jun 19, 2026
- 1
- 1.5
- 2
- 2.5
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The Correct Option is C
Solution and Explanation
Step 1: Use conservation of mechanical energy.
For motion under gravity: \[ \frac{1}{2}mV_0^2 - \frac{GMm}{R} = -\frac{GMm}{\frac{4R}{3}} \]
Step 2: Simplify potential energies.
\[ \frac{1}{2}mV_0^2 = GMm\left(\frac{1}{R} - \frac{3}{4R}\right) \]
Step 3: Simplify bracket.
\[ \frac{1}{R} - \frac{3}{4R} = \frac{1}{4R} \]
So, \[ \frac{1}{2}mV_0^2 = \frac{GMm}{4R} \]
Step 4: Solve for \(V_0^2\).
\[ V_0^2 = \frac{GM}{2R} \]
Step 5: Use escape velocity.
\[ V_E^2 = \frac{2GM}{R} \]
Step 6: Find ratio.
\[ \frac{V_E^2}{V_0^2} = \frac{\frac{2GM}{R}}{\frac{GM}{2R}} = 4 \Rightarrow \frac{V_E}{V_0} = 2 \]
Final Answer: \[ \boxed{2} \]
For motion under gravity: \[ \frac{1}{2}mV_0^2 - \frac{GMm}{R} = -\frac{GMm}{\frac{4R}{3}} \]
Step 2: Simplify potential energies.
\[ \frac{1}{2}mV_0^2 = GMm\left(\frac{1}{R} - \frac{3}{4R}\right) \]
Step 3: Simplify bracket.
\[ \frac{1}{R} - \frac{3}{4R} = \frac{1}{4R} \]
So, \[ \frac{1}{2}mV_0^2 = \frac{GMm}{4R} \]
Step 4: Solve for \(V_0^2\).
\[ V_0^2 = \frac{GM}{2R} \]
Step 5: Use escape velocity.
\[ V_E^2 = \frac{2GM}{R} \]
Step 6: Find ratio.
\[ \frac{V_E^2}{V_0^2} = \frac{\frac{2GM}{R}}{\frac{GM}{2R}} = 4 \Rightarrow \frac{V_E}{V_0} = 2 \]
Final Answer: \[ \boxed{2} \]
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