The heightof the point vertically above the earth surface of radius at which acceleration due to gravity become 1% of its value at the surface is (R is the radius of the earth)
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1 ℅ of the acceleration due to gravity at the surface is 0.098
For earth this calculation is possible as it Newtonian physics deals with it
gR^2 = 0.098r^2
Where r^2 is the place where the accelacceleration becomes 1 ℅ of actual acceleration on the surface of the earth
9.8R^2 = 0.098r^2
100R^2 = r^2
Taking square root on both the sides it gives
10R = r
And hence you have asked from the surface of the earth
10R - R = 9R
For earth this calculation is possible as it Newtonian physics deals with it
gR^2 = 0.098r^2
Where r^2 is the place where the accelacceleration becomes 1 ℅ of actual acceleration on the surface of the earth
9.8R^2 = 0.098r^2
100R^2 = r^2
Taking square root on both the sides it gives
10R = r
And hence you have asked from the surface of the earth
10R - R = 9R
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