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svetoff [14.1K]
3 years ago
13

When a falling meteoroid is at a distance above the Earth's surface of 3.40 times the Earth's radius, what is its acceleration d

ue to the Earth's gravitation
Physics
1 answer:
Alchen [17]3 years ago
6 0

Answer:

g = 0.85 ms^{-2}

Explanation:

g = \frac{GM}{h^{2} }

were; g is the acceleration due to Earth's gravity, G is Newton's gravitation constant (6.674 x 10^{-11} Nm^{2}kg^{-2}), M is the mass of the earth (5.972 x 10^{24} kg), and h is the distance of meteoroid to the earth.

h = 3.40 x R

  = 3.40 x 6371 km

h = 21661.4 km

  = 21661400 m

Thus,

g = \frac{6.674*10^{-11}*5.972*10^{24}  }{(21661400)^{2} }

  = \frac{3.9857 *10^{14} }{4.6922*10^{14} }

  = 0.84944

g = 0.85 ms^{-2}

The acceleration due to the Earth's gravitation is 0.85 ms^{-2}.

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