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?

Respuesta :

Answer:

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

Explanation:

We can use the gravitational force equation:

[tex]F_{g}=G\frac{mM_{e}}{R^{2}}[/tex]

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

[tex]ma=G\frac{mM_{e}}{R^{2}}[/tex]

[tex]a=G\frac{M_{e}}{R^{2}}[/tex]

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

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

[tex]a=6.67*10^{-11}\frac{5.972*10^{24}}{(2*6.377*10^{6})^{2}}[/tex]  

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

I hope it helps you!