Physics, asked by surbhiarora4701, 9 months ago

Calculate acceleration due to gravity at a high of 300 km from the surface of the earth (m= 5.98×10^24kg, r=6400km

Answers

Answered by aditya012
2

Answer:

The acceleration due to gravity at a height of 300 km above the surface of the earth is 8.9 m/s^2

Explanation:

g = GM/R^2

Here

M = mass of the earth = 5.98 x 10^24 kg

R = radius of the Earth + the extra distance = 6400 + 300

= 6700 km = 6700000 m = 6.7 x 10^6 m

G = gravitational constant = 6.7 x 10^-11 N m^2/kg^2

=> g = 6.7 x 10^-11 x 5.98 x 10^24 / 6.7 x 6.7 x 10^12

=> g = 5.98 x 10^13 / 6.7 x 10^12

=> g = 0.89 x 10

=> g = 8.9 m/s^2 ( approx)

EXTRA:

Decrease %

decrease = 0.9

original = 9.8

Decrease % = 0.9/9.8 x 100

= 9.18 %

The acceleration due to gravity decrease by roughly 9.18% 300 km above the surface of the earth.

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