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yarga [219]
2 years ago
14

At what height from the surface of earth will the value of 'g' reduced to 64% from the value at the surface? Radius of earth = 6

400 Km​
Physics
1 answer:
Sonbull [250]2 years ago
8 0

Answer: 1600 km from the surface

Explanation:

E = \frac{GM}{R^{2} }\\g1 = \frac{GM}{R1^{2} } \\R1 = 6400 km\\g2 = 0.64g1\\\frac{g1}{g2} = \frac{\frac{GM}{R1^{2} } }{\frac{GM}{R2^{2} } } \\\frac{g1}{0.64g1} = \frac{R2^{2}}{R1^{2} }\\\frac{1}{0.64} = \frac{R2^{2}}{6400^{2} }\\\\\frac{1}{0.8} = \frac{R2}{6400}\\\\R2 = \frac{6400}{0.8}\\R2 = 8000 km\\

Therefore, 8000 km - 6400 km = 1600 km from the surface

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This question is incomplete, the complete question is;

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