Question
Earths mass is 6 x 10^24kg and its radius is 6.38 x 10^6m
(6380 km ). Use the inverse-square law to show that in space-shuttle territory, 200 km above the earths surface the force of gravity on a shuttle is about 94% of that at the earths surface.
(6380 km ). Use the inverse-square law to show that in space-shuttle territory, 200 km above the earths surface the force of gravity on a shuttle is about 94% of that at the earths surface.
Answers
Let the acceleration of gravity at 200 km altitude be g' and the value at sea level be g. Then
g'/g = (6380/6580)^2 = 0.94013
The rounds off to 94%, to the nearest 1%.
g'/g = (6380/6580)^2 = 0.94013
The rounds off to 94%, to the nearest 1%.
Related Questions
Find the gravitational force between earth and the sun.
I am using the formula Fg = (G*m1*m2)/(r^...
What is the value of g for space-shuttle territory , about 200km above the Earth's surface. Data: Ea...
The shuttle is orbiting the earth at a distance of 500km above the earths surface. Assumingan astron...