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Question

If the value of g at Earth's surface is 10 m/s², what is its value at a depth of R/2 (where R is Earth's radius)?

RankGuru · Accepted Answer

1. **Gravity Variation with Depth**: $g_d = g \left( 1 - \frac{d}{R} ight)$.
2. **Identify Depth**: $d = R/2$.
3. **Calculation**:
   - $g_d = 10 	imes (1 - \frac{R/2}{R})$.
   - $g_d = 10 	imes (1 - 0.5)$.
4. **Result**: $g_d = 5$ m/s$^2$.
5. **Context**: Gravitation inside the Earth decreases linearly with distance from the center, reaching zero at $d=R$ (the Earth's core).

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