This is an interesting problem. You will have to integrate the force of gravity from each dmass. Fortuantly, symettry makes it easier, as you can use symettry to just integrate one component of the force (the other side will cancel that component).
Because of symettry, make the box shape a wire of mass M and length 40km.
MassM is rho*volume.
So mass per unit length= rho*volume/40km
Now draw a triangle, one side the perpendiular of length 15km, the base of x (x goes from zero to 20km), and the hypotenuse of length=sqrt(225E3+x^2)
Now the angle theta at the apex of the triangle has a cosine of 15km/x. Notice that we can use the cosine of theta, as that is the force of attraction perpendicular to the rod.
Forcetotal=2*Gm INT dM/ sqrt(225E6+x^2) * cosTheta
but cosTheta= 15E3/sqrt(225E6+x^2)
and dM= rho*volume dx/40E3
so all that reduced is integrable.
Rember F/m= acceleartion which is equal to v^2/r and v= 2PIr/T so you can solve for T
Work it out for the sphere, just assume a point mass at 15km, see if it is different.
the near earth asteroid Rendezvous (NEAR), after travelling 2.1 billion km, is meant to orbit the asteroid Eros at a height of about 15 km from the asteroidal center. Eros is roughly box-shaped, with dimensions 40km x 6km x 6km. Assume eros has a density (mass/ volume) of about 2.3 x 10^3 kg/m^3.
a) what will be the period of NEAR as it orbits Eros?
b) If eros were a sphere witht he same mass and density, waht would be its radius.
c)what would g be at the surface of this spherical Eros?
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