Question

If you were at a height of 16197 above see level , what is the acceleration due to gravity objects allowed to fall freely at this altitude

Answer 1

acceleration due to gravity at 16197 km above the ground = 2.76
`(m)/(s^(2))`

Step-by-step explanation:

force of gravity two bodies
`(GMm)/(r^(2))`

where G is the gravitational constant, M is the mass of earth, m the mass of other body, and r the distance of separation.

when the body is on the surface of earth acceleration due to gravity will be

`(GM)/(r^(2))` = g

when the body is at a height h above the surface of the ground acceleration due to gravity will be
`(GM)/((r+h)^(2))`

`(GM)/((r+h)^(2))` =
`(GM)/(r^(2)(1+(h)/(r)))`

`(GM)/(r^(2)(1+(h)/(r)))` =
`(g)/(1+(h)/(r))`

we know that g = 9.8
`(m)/(s^(2))`

therefore acceleration due to gravity =
`(9.8)/(1+(16197)/(6378))`

acceleration due to gravity =
`(9.8)/(1+(16197)/(6378))`

=2.76
`(m)/(s^(2))`