Question

An astronaut goes out for a "space-walk" at a distance above the earth equal to twice the radius of the earth. What is her acceleration due to gravity?

Answer 1

Answer:

Acceleration due to gravity will be $a=2.45 m/s^{2}$.

Explanation:

We can use the gravitational force equation:

$F_{g}=G\frac{mM_{e}}{R^{2}}$

The F is equal to the weight of the astronaut, so we will have:

$ma=G\frac{mM_{e}}{R^{2}}$

$a=G\frac{M_{e}}{R^{2}}$

• M(e) is the mass of the earth $M_{e}=5.972*10^{24}kg$
• R is the radius of the earth $R=6.377*10^{6}m$
• G is the gravitational constant $G=6.67*10^{-11}m^{3}kg^{-1}s^{-2}$

But the distance between the astronaut and the center of the earth is 2R, then we have:

$a=6.67*10^{-11}\frac{5.972*10^{24}}{(2*6.377*10^{6})^{2}}$  

Therefore the acceleration due to gravity will be $a=2.45 m/s^{2}$.

I hope it helps you!