The acceleration due to gravity g at height h above Earth's surface (radius R) is approximately
A{'text': 'g × (R + h)', 'label': 'A'}
B{'text': 'g × (1 + 2h/R)', 'label': 'B'}
C{'text': 'g × (1 − 2h/R)', 'label': 'C'}
D{'text': 'g × (R − h)', 'label': 'D'}
Answer & Solution
Correct answer: C. {'text': 'g × (1 − 2h/R)', 'label': 'C'}
1. g_h = g × R² / (R + h)².
2. For h ≪ R, expand: (1 + h/R)⁻² ≈ 1 − 2h/R.
3. So g_h ≈ g × (1 − 2h/R) to first order.
4. This is the fractional decrease in g as we rise above Earth's surface.
_Source: NCERT Class 11 Physics, Ch 7 "Gravitation", §7.4_
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