Question

at what height above the surface of the earth acceleration due to gravity reduced by 1 % answer =32km

Answer 1

3 answers · Physics 

Answers

from the equation F = G m1 m2 / r^2 
we see that the force of gravity is inversely proportional to the square of the distance between the centers of the objects attracting each other. 
Since F also equals m a it is a short step to see that g is inversely proportional to r^2 
so 
g_h / g = [R / (R+h)]^2 
you can put in g and the radius of the Earth then solve for h 
but this leads to a quadratic equation. 
An easier way is to solve for (R+h) first then subtract R. 

From the given g_h = 0.99 g 
so 
0.99 g/ g = [R/(R+h)]^2 
g's cancel and take sqrt of both sides 
0.9945 = R/(R+h) 
(R+h) = R / 0.9945 = 6.371E6 m / 0.9945 = 6.4062E6 m 
h = 6.40+3E6 - 6.371E6 = 3.52E4 m or 35.2 km above the surface of the Earth 

Answer 2

from the equation F = G m1 m2 / r^2

we see that the force of gravity is inversely proportional to the square of the distance between the centers of the objects attracting each other.

Since F also equals m a it is a short step to see that g is inversely proportional to r^2

so

g_h / g = [R / (R+h)]^2

you can put in g and the radius of the Earth then solve for h

but this leads to a quadratic equation.

An easier way is to solve for (R+h) first then subtract R.

From the given g_h = 0.99 g

so

0.99 g/ g = [R/(R+h)]^2

g's cancel and take sqrt of both sides

0.9945 = R/(R+h)

(R+h) = R / 0.9945 = 6.371E6 m / 0.9945 = 6.4062E6 m

h = 6.40+3E6 - 6.371E6 = 3.52E4 m or 35.2 km above the surface of the Earth