Hey Guys. I\'m having a lot of difficulty with this one. Even looking at other e
ID: 1895985 • Letter: H
Question
Hey Guys. I'm having a lot of difficulty with this one. Even looking at other examples I have problems. Could someone work this out, and not only give the answers, but a good explanation and step-by-step process with this. Something else that would be helpful is if you could write down a list of the equations necessary to solve.Image: http://www.webassign.net/serpse8/13-p-027.gif
A spacecraft in the shape of a long cylinder has a length of 100 m, and its mass with occupants is 1 280 kg. It has strayed too close to a black hole having a mass 93 times that of the Sun. The nose of the spacecraft points toward the black hole, and the distance between the nose and the center of the black hole is 10.0 km.
(a) Determine the total force on the spacecraft.
N
(b) What is the difference in the gravitational fields acting on the occupants in the nose of the ship and on those in the rear of the ship, farthest from the black hole? (This difference in acceleration grows rapidly as the ship approaches the black hole. It puts the body of the ship under extreme tension and eventually tears it apart.)
N/kg
Explanation / Answer
a)
total force = F = GMm/r2
r=distance between two centers of mass=10,000+50=10050m
M=mass of the black hole=93 times that of the Sun=93*(1.99e30)
>>> M = 185.07e30 Kg
>>>> F = GMm/r2
= (6.67e-11)*(185.07e30)*(1280)/(10050*10050)
= 1.56*10^17 N
-----------------------------------------------
b)
gravitational fields acting on the occupants in the nose of the ship = Fn
Fn = GM/r2 = (6.67e-11)*(185.07e30)/(10000*10000) = 1.2344*10^14 N/kg
gravitational fields acting on the occupants in the nose of the ship = Fr
Fr = GM/r2 = (6.67e-11)*(185.07e30)/(10100*10100) = 1.2101*10^14 N/kg
the difference in the gravitational fields = Fr - Fn = (1.234-1.21)*10^14 = 2.43*10^12 N/kg