Question

What is the gravitational force on a 70kg that is 6.38x10^6m above the earths surface

Answer 1

171.5 N

Step-by-step explanation:

The gravitational force on an object due to the Earth is given by

`F=mg`

where

m is the mass of the object

g is the acceleration due to gravity

The acceleration due to gravity at a certain height h above the Earth is given by

`g=(GM)/((R+h)^2)`

where:

G is the gravitational constant

`M=5.98\cdot 10^(24) kg` is the Earth's mass

`R=6.37\cdot 10^6 m` is the Earth's radius

Here,

`h=6.38\cdot 10^6 m`

So the acceleration due to gravity is

`g=((6.67\cdot 10^(-11))(5.98\cdot 10^(24)))/((6.37\cdot 10^6 + 6.38\cdot 10^6)^2)=2.45 m/s^2`

We know that the mass of the object is

m = 70 kg

So, the gravitational force on it is

`F=mg=(70)(2.45)=171.5 N`