Question:
The escape speed of an object on the surface of the Earth is \(V\). If the object is thrown out with speed \(4V\) from the surface of the Earth, find the speed of the object far away from the Earth.
The escape speed of an object on the surface of the Earth is \(V\). If the object is thrown out with speed \(4V\) from the surface of the Earth, find the speed of the object far away from the Earth.
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Use energy conservation for escape problems: \(\frac{1}{2} m v^2 - \frac{GMm}{r} = \text{constant}\) to find final speed at infinity.
Updated On: Jul 18, 2026
- \(3V\)
- \(\sqrt{15} V\)
- \(2.5V\)
- \(\sqrt{8} V\)
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The Correct Option is B
Solution and Explanation
Step 1: Recall escape velocity relation.
Escape velocity \(V\) satisfies \(\frac{1}{2} m V^2 = \frac{GMm}{R}\) where \(R\) is radius of Earth and \(GM\) is gravitational constant times mass of Earth.
Step 2: Apply energy conservation.
Initial kinetic energy with speed \(4V\) and potential energy at surface:
\[ \frac{1}{2} m (4V)^2 - \frac{GMm}{R} = \frac{1}{2} m v_\infty^2 \]
where \(v_\infty\) is final speed far away.
Step 3: Express GM/R in terms of V.
\[ \frac{GM}{R} = \frac{1}{2} V^2 \implies \frac{GMm}{R} = \frac{1}{2} m V^2 \]
Step 4: Substitute values.
\[ \frac{1}{2} m (16 V^2) - \frac{1}{2} m V^2 = \frac{1}{2} m v_\infty^2 \]
\[ \frac{1}{2} m (15 V^2) = \frac{1}{2} m v_\infty^2 \]
Step 5: Solve for \(v_\infty\).
\[ v_\infty = \sqrt{15} V \]
Step 6: Final conclusion.
Hence, the speed of the object far away from the Earth is:
\[ \boxed{\sqrt{15} V} \]
Escape velocity \(V\) satisfies \(\frac{1}{2} m V^2 = \frac{GMm}{R}\) where \(R\) is radius of Earth and \(GM\) is gravitational constant times mass of Earth.
Step 2: Apply energy conservation.
Initial kinetic energy with speed \(4V\) and potential energy at surface:
\[ \frac{1}{2} m (4V)^2 - \frac{GMm}{R} = \frac{1}{2} m v_\infty^2 \]
where \(v_\infty\) is final speed far away.
Step 3: Express GM/R in terms of V.
\[ \frac{GM}{R} = \frac{1}{2} V^2 \implies \frac{GMm}{R} = \frac{1}{2} m V^2 \]
Step 4: Substitute values.
\[ \frac{1}{2} m (16 V^2) - \frac{1}{2} m V^2 = \frac{1}{2} m v_\infty^2 \]
\[ \frac{1}{2} m (15 V^2) = \frac{1}{2} m v_\infty^2 \]
Step 5: Solve for \(v_\infty\).
\[ v_\infty = \sqrt{15} V \]
Step 6: Final conclusion.
Hence, the speed of the object far away from the Earth is:
\[ \boxed{\sqrt{15} V} \]
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