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:

• A. \( 5:7 \)
• B. \( 9:7 \)
• C. \( 3:7 \)
• D. \( 11:7 \)

Hint:

For vertical projection at a fraction \( f \) of escape velocity, the ratio of maximum height to radius is \( \frac{f^2}{1-f^2} \).

Answer 1

The Correct Option is B

Step 1: Recall escape speed.
The escape speed from the Earth's surface is $v_e = \sqrt{\dfrac{2GM}{R}}$, so $v_e^2 = \dfrac{2GM}{R}$. We will use energy conservation to find the height reached.
Step 2: Write the launch speed.
The body is thrown at $75\%$ of escape speed, so $v = \dfrac{3}{4}v_e$. Squaring, \[ v^2 = \frac{9}{16}v_e^2 = \frac{9}{16}\cdot\frac{2GM}{R} = \frac{9GM}{8R}. \]
Step 3: Total energy at the surface.
Energy is kinetic plus gravitational potential: \[ E = \frac{1}{2}mv^2 - \frac{GMm}{R} = \frac{9GMm}{16R} - \frac{GMm}{R} = -\frac{7GMm}{16R}. \]
Step 4: Energy at the highest point.
At the top the speed is zero, and the distance from Earth's centre is $R+h$. So the energy there is purely potential: \[ E = -\frac{GMm}{R+h}. \]
Step 5: Equate the two energies.
Energy is conserved, so \[ -\frac{7GMm}{16R} = -\frac{GMm}{R+h} \;\Rightarrow\; \frac{7}{16R} = \frac{1}{R+h}. \] Cross-multiplying, $7(R+h) = 16R$.
Step 6: Solve the ratio.
Then $7h = 16R - 7R = 9R$, giving $\dfrac{h}{R} = \dfrac{9}{7}$. So the maximum height to Earth radius ratio is $9:7$.
\[ \boxed{9:7} \]