Question

2. At what distance above the surface of the earth is the acceleration due to the earth's gravity 0.835 m/s2 if the acceleration due to gravity at the surface has magnitude 9.80m/s2?

______ m

Can you finish this problem before 12 pm central time cause around that time is when the assignment is due. Thank you for your time, have a great day!

Answer 1

here,

let the distance from the surface of earth is x

as g = go/(1 + h/R)^2

where R is the radius of earth

0.835 = 9.8/(1 + h/6371)^2

solving for h

x = 15422.2 km = 1.54 *10^7 m

the distance from the surface of earth is 1.54 *10⁷ m

Answer 2

SOLUTION :

Let that height be h  m from the earth’s surface at which altitude, g’ = 0.835 m/s^2

g at the earth’s surface is = 9.8 m/s^2.

As we go up, value of acceleration due to gravity decreases.

Value of g’ at height h is given by the following formula:

g’ = g (1 - 2h / R) 

Where,

g’ = acceleration due to gravity at height h = 0.835  m/s^2 (given in the problem)

g = acceleration due to gravity at earth’s surface = 9.8 m/s^2

h = height at which g’ = 0.835 m/s^2. (To be determined)

R = radius of the earth = 6371 km = 6.371*10^6 m

So, as per above formula :

0.835 = 9.8( 1 - 2 h / (6.371 * 10^6) )

=> h = (1 - 0.835/9.8) *  (6.371 * 10^6) / 2 

=> h = 2 914 082.40 m = 2914.082 km

=> h = height at which g’ is 0.835 m/s^2 = 2914.082 km  approx.  (ANSWER).