Question

A rocket is fired vertically from the surface of the earth with quarter the escape speed. If R is radius of the earth, the maximum altitude reached by the rocket is:

• A. R/5
• B. R/3
• C. R/15
• D. R/14

Hint:

Energy conservation is the key! Use $v_{escape} = \sqrt{2gR}$.

Answer 1

The Correct Option is C

Step 1: Use energy conservation.
The rocket only feels gravity once it leaves the surface, so its total energy (kinetic plus gravitational potential) stays the same from launch to the highest point, where its speed is zero.

Step 2: Recall the escape speed.
The escape speed from Earth is \[ v_e = \sqrt{\frac{2GM}{R}} \] The rocket is fired at one quarter of this, so $v = \dfrac{v_e}{4}$ and \[ v^2 = \frac{v_e^2}{16} = \frac{1}{16}\cdot\frac{2GM}{R} = \frac{GM}{8R} \]

Step 3: Write the energy equation.
Let $H$ be the maximum height. Energy at the surface equals energy at the top: \[ \tfrac{1}{2}mv^2 - \frac{GMm}{R} = -\frac{GMm}{R+H} \]

Step 4: Substitute $v^2$.
\[ \tfrac{1}{2}\cdot\frac{GM}{8R} - \frac{GM}{R} = -\frac{GM}{R+H} \] \[ \frac{GM}{16R} - \frac{GM}{R} = -\frac{GM}{R+H} \] Cancel $GM$ and combine the left side: \[ \frac{1}{16R} - \frac{1}{R} = \frac{1 - 16}{16R} = -\frac{15}{16R} \]

Step 5: Solve for $H$.
\[ -\frac{15}{16R} = -\frac{1}{R+H} \;\Rightarrow\; R+H = \frac{16R}{15} \] \[ H = \frac{16R}{15} - R = \frac{16R - 15R}{15} = \frac{R}{15} \]

Step 6: Conclusion.
The maximum altitude reached is $\dfrac{R}{15}$. \[ \boxed{\dfrac{R}{15}} \]