Question:
If a body is thrown vertically upwards from the surface of the earth with a speed equal to 75% of the escape speed from the surface of the earth, then the ratio of the maximum height reached by the body and the radius of the earth is:
If a body is thrown vertically upwards from the surface of the earth with a speed equal to 75% of the escape speed from the surface of the earth, then the ratio of the maximum height reached by the body and the radius of the earth is:
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For vertical projection at a fraction \( f \) of escape velocity, the ratio of maximum height to radius is \( \frac{f^2}{1-f^2} \).
Updated On: Jun 9, 2026
- \( 5:7 \)
- \( 9:7 \)
- \( 3:7 \)
- \( 11:7 \)
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The Correct Option is B
Solution and Explanation
Concept:
We utilize the principle of Conservation of Energy in a gravitational field. The total energy (kinetic + potential) at the surface is equal to the total energy at the maximum height (where kinetic energy is zero). Escape velocity \(v_e = \sqrt{\frac{2GM}{R}}\).
Step 1: Define initial energy.
The velocity of projection is \( v = 0.75 v_e = \frac{3}{4} \sqrt{\frac{2GM}{R}} \). $$ E_{total} = -\frac{GMm}{R} + \frac{1}{2}m v^2 = -\frac{GMm}{R} + \frac{1}{2}m \left( \frac{9}{16} \cdot \frac{2GM}{R} \right) $$ $$ E_{total} = -\frac{GMm}{R} + \frac{9GMm}{16R} = -\frac{7GMm}{16R} $$
Step 2: Define energy at maximum height \( H \).
At the maximum height \( (R+h) \), velocity is zero: $$ E_{final} = -\frac{GMm}{R+h} $$
Step 3: Equate energies and solve for \( h \).
$$ -\frac{7GMm}{16R} = -\frac{GMm}{R+h} $$ $$ \frac{7}{16R} = \frac{1}{R+h} \implies 7(R+h) = 16R $$ $$ 7R + 7h = 16R \implies 7h = 9R \implies \frac{h}{R} = \frac{9}{7} $$ $$\boxed{9:7}$$
Step 1: Define initial energy.
The velocity of projection is \( v = 0.75 v_e = \frac{3}{4} \sqrt{\frac{2GM}{R}} \). $$ E_{total} = -\frac{GMm}{R} + \frac{1}{2}m v^2 = -\frac{GMm}{R} + \frac{1}{2}m \left( \frac{9}{16} \cdot \frac{2GM}{R} \right) $$ $$ E_{total} = -\frac{GMm}{R} + \frac{9GMm}{16R} = -\frac{7GMm}{16R} $$
Step 2: Define energy at maximum height \( H \).
At the maximum height \( (R+h) \), velocity is zero: $$ E_{final} = -\frac{GMm}{R+h} $$
Step 3: Equate energies and solve for \( h \).
$$ -\frac{7GMm}{16R} = -\frac{GMm}{R+h} $$ $$ \frac{7}{16R} = \frac{1}{R+h} \implies 7(R+h) = 16R $$ $$ 7R + 7h = 16R \implies 7h = 9R \implies \frac{h}{R} = \frac{9}{7} $$ $$\boxed{9:7}$$
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