Question

The escape speed from a distant planet is 47 km/s. If the acceleration of gravity on the planet surface is 15.0 m/s2, what is the radius of the planet?

Answer 1

Answer 2

Given:

• Escape speed, $v_e=47\text{km/s}=47,\text{\hspace{0.17em}⁣}000\text{m/s}$

• Surface gravity, $g=15.0{\text{m/s}}^2$

Formula for escape speed:

$$v_e=\sqrt{\frac{2GM}{R}}$$

Formula for surface gravity:

$$g=\frac{GM}{R^2}$$

Combine the formulas:

1. Solve for $GM$ from gravity:

   $$GM=g{R}^2$$

2. Substitute into escape speed:

   $$v_e=\sqrt{\frac{2g{R}^2}{R}}=\sqrt{2gR}$$

3. Solve for $R$:

   $$R=\frac{v_e^2}{2g}=\frac{(47,\text{\hspace{0.17em}⁣}000{)}^2}{2\times 15.0}$$$$R=\frac{2,\text{\hspace{0.17em}⁣}209,\text{\hspace{0.17em}⁣}000,\text{\hspace{0.17em}⁣}000}{30}=73,\text{\hspace{0.17em}⁣}633,\text{\hspace{0.17em}⁣}333\text{m}$$$$R\approx 73,\text{\hspace{0.17em}⁣}633\text{km}$$

Final Answer:
The planet's radius is 73,633 km