acceleration due to gravity

Question

At what height above the Earth's surface will the acceleration due to gravity be half of its value at the surface? (R = radius of Earth)

RankGuru · Accepted Answer

1. **Gravity Formula**: Acceleration at surface $g = \frac{GM}{R^2}$. Acceleration at height $h$ is $g_h = \frac{GM}{(R+h)^2}$.
2. **Identify Condition**: $g_h = \frac{g}{2}$.
3. **Set up Equation**:
   - $\frac{GM}{(R+h)^2} = \frac{1}{2} \frac{GM}{R^2}$
   - $\frac{1}{(R+h)^2} = \frac{1}{2R^2}$.
4. **Solve for $h$**:
   - Take square root: $R + h = R \sqrt{2}$.
   - $h = R \sqrt{2} - R = R(\sqrt{2} - 1)$.
5. **Approximate Value**: Since $\sqrt{2} \approx 1.414$, $h \approx 0.414 R$. Note: Do NOT use $g_h = g(1 - \frac{2h}{R})$ here as $h$ is not very small compared to $R$.

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