# 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...

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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.

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