R=6.37X10^6G*M/(R+366.1X10^3)=gGravity Force on 1.09 kg Sphere in Space Shuttle

  • Thread starter Thread starter talaroue
  • Start date Start date
  • Tags Tags
    Gravity Theory
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
4 replies · 3K views
talaroue
Messages
302
Reaction score
0

Homework Statement



The space shuttle orbits 366.1 km above the surface of the earth. What is the gravitational force on a 1.09 kg sphere inside the space shuttle?

Homework Equations



U=Gm1m2/Re
U=-W
W=m(freefall)

The Attempt at a Solution



I am starting to understand the theory behind gravity now, am I setting this problem up correctly? I am making U=-W and then solving for the free fall so I come up with ...

G*M/(radius of earth+ distance of how far it is from earth)=g

G=6.67X10^-11
M=5.98X10^24
 
Physics news on Phys.org
What is the formula of the gravitational force between two objects?
 
I assume that by W you mean work (e.g. work done against the gravitational force to lift the sphere from the center of the Earth to its current position). In that case, W is not given by m, but by m g h where h = radius of Earth + distance above the surface. It is probably easier, however, to go from the potential energy directly to the force through F = - U', where U denotes differentiation with respect to the distance, giving
F = G M m / r^2
where M, m are the mass of the Earth and sphere and r is the distance between the sphere and the center of the Earth (= radius of Earth + distance above the surface).

You can also use the first two formulas you gave complemented with W = m g r as I explained above to derive an expression for g first, and then use F = m g for the gravitational force. That will give you the same result.
 
So all I have to do is ues F=G*M*m/r^2...? Was I just over thinking things?
Where F is the gravitational force?