Gravitational Forces on Ship Stranded Near Black Hole

AI Thread Summary
A ship shaped like a cylinder, with a mass of 1000 kg, is 10 km from a black hole with a mass 100 times that of the sun. The total gravitational force on the ship is calculated to be approximately 1.3 x 10^17 N. For the occupants, the gravitational field differs between the front and rear of the ship due to their varying distances from the black hole, with the front at 10 km and the rear at 10.1 km. The difference in gravitational fields highlights the "spaghettification" effect experienced by the occupants. Calculating the total force should consider the center of mass, while the gravitational field difference is determined solely by the black hole's mass and the respective distances.
rpcarroll
Messages
6
Reaction score
0

Homework Statement


A ship in the shape of a cylinder has length 100m, and mass 1000kg with occupants. It has strayed too close to a black hole with mass 100 times greater than the sun (100*2.0*10^30). The nose of the ship is pointed towards the black hole and is 10km from the center of the black hole.
(a) What is the total force on the ship?
(b) What is the difference in the gravitational fields acting on the occupants in the front of the ship compared to those in the rear (furthest from the hole)?


Homework Equations


Gravitational Field= (Fg/m)
Fg=G * [(m1*m2)/r2]


The Attempt at a Solution



(a) Fg= [1000kg*(100*2.0*1030]/100002m=1.3*1017N

Not sure about b, do I need to calculate Fg at the two different distances (front and rear of the ship)?
 
Physics news on Phys.org
rpcarroll said:

Homework Statement


A ship in the shape of a cylinder has length 100m, and mass 1000kg with occupants. It has strayed too close to a black hole with mass 100 times greater than the sun (100*2.0*10^30). The nose of the ship is pointed towards the black hole and is 10km from the center of the black hole.
(a) What is the total force on the ship?
(b) What is the difference in the gravitational fields acting on the occupants in the front of the ship compared to those in the rear (furthest from the hole)?


Homework Equations


Gravitational Field= (Fg/m)
Fg=G * [(m1*m2)/r2]


The Attempt at a Solution



(a) Fg= [1000kg*(100*2.0*1030]/100002m=1.3*1017N

Not sure about b, do I need to calculate Fg at the two different distances (front and rear of the ship)?

I'm assuming that's what they want to know. :)
 
Pretty much that's the way to do it looks like to me.

Those in front are at 10 km and those in the back are at 10.1 km.
 
The idea of question b is for you to figure out how large the "spaghettification" effect is.
You'll know what it is once you've calculated it :-)
 
So when they ask for the 'total force on the ship', is it just measured at 10km and not, say, 10.05 (I think that would be the center of mass of the cylinder shaped ship)?
 
Last edited:
rpcarroll said:
So when they ask for the 'total force on the ship', is it just measured at 10km and not, say, 10.05 (I think that would be the center of mass of the cylinder shaped ship)?

With part a) yes I think you can use the center of mass to calculate the total force. (10.05 km would be good, though at these numbers the precision within 1/2 % is not exactly that important.)

For part b) they are looking for the difference in the gravitational field.

That only involves the mass of the black hole, not the whole ship. That's given more simply by GM/r
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top