At what height will the acceleration due to gravity be 6.0 m/s^2 and what is the formula for this given the following:
The acceleration due to gravity for a satellite orbiting 7000 km above the centre of Earth is 8.2 m/s^2.
The force of attraction due to gravity between an object of mass m and the Earth of mass Me is G*m^Me/ r^2. This, in terms of the acceleration due to gravity of the object is m*a
Equating the two m*g = G*m^Me/ r^2
=> g = k/r^2, where k is a constant as G*Me is a constant.
At a height of 7000 km the acceleration due to gravity is 8.2 m/s^2
8.2 = k/ 7000^2
=> k = 8.2 * 7000^2
The formula for the relationship between height and the acceleration due to gravity is g = (8.2*7000^2) / r^2
When the acceleration is 6 we have :
6 = (8.2*7000^2) / r^2
=> r = sqrt [(8.2*7000^2) / 6]
=> r = 8417 km.
The required formula is g = (8.2*7000^2) / r^2 and the required height is 8417 km.