Calculate the effective value of the acceleration of gravity at 6100m above the earth's surface?

Answer 1

Assuming all objects fall at the same pace at the same altitude, using Newton's formula of Gravity, F = G*m1*m2/r2.

G is the global gravity constant, m1 = 5.9736 × 1024 kg, m2 = 1kg (because it doesn't matter, all objects fall at the same pace, 1 is easier to calculate), and r2 = (earth's radius + 6100m)^2

After you put all those numbers, you get F =~ 9.8N. Since our object is 1kg of mass, that's also the acceleration (according to newton's 2nd law)

Answer 2

At a height of 6100m above the Earth's surface, the effective value of the acceleration of gravity would be slightly less than the value at the Earth's surface due to the increase in distance from the Earth's center. This decrease in acceleration is given by the formula: ( g' = g \left( \frac{R}{R+h} \right)^2 ), where g is the acceleration of gravity on the surface of the Earth (approximately 9.81 m/s^2) and R is the radius of the Earth (approximately 6371 km). By applying this formula, you can find the effective value of the acceleration of gravity at 6100m above the Earth's surface.