Force required to keep objects in orbit.

In summary: F = Gm_{1}m_{2}/r^2 = m_{2}v^2/r = maThere are two forces here. The gravity force, Gm_{1}m_{2}/r^2, and the centripetal force, m_{2}v^2/r. The mass m_{2} is the same in both cases, but the other mass, m_{1}, in the gravity force is not the same as the mass in the centripetal force.What you want to do is find the velocity that makes the two forces equal.You can do this algebraically, since you have two equations with two unknowns (G and v).You could also plug
  • #1
the7joker7
113
0

Homework Statement




Neutron stars are extremely dense objects that are formed from the remnants of supernova explosions. Many rotate very rapidly. Suppose that the mass of a certain spherical neutron star is twice the mass of the Sun and its radius is 10.0 km. Determine the greatest possible angular speed it can have for the matter at the surface of the star on its equator to be just held in orbit by the gravitational force.

Homework Equations



F = G*[tex]\frac{v^{2}}{r}[/tex]

The Attempt at a Solution



The way I see it, r = 10,000 meters and mass = 3.977 *10[tex]^{30}[/tex] kilograms. So I used the equation above. to get...

F = (6.67*10^-11)*[tex]\frac{v^{2}}{10000}[/tex]

Now, I think I'm supposed to find out what F should be, and solve for V, but I'm not sure what I'm supposed to make F be.
 
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  • #2
I don't know why you tossed the gravitational constant into the equation for centripetal force

You have two forces going on here, centripetal and gravity, but what IS actually causing the centripetal force? What is holding the mass in place?

Anyways this proceeds extremely similarly to the question "how fast is a satellite at such and such orbit around the Earth traveling?" Note that I didn't give the mass of the satellite. You find that speed, and that is THE speed for that orbit. If you slow it down, it falls into a closer orbit, if you speed it up...
 
  • #3
the7joker7 said:

Homework Equations



F = G*[tex]\frac{v^{2}}{r}[/tex]
Hi!

You need two relevant equations: one for the acceleration (which you've got), and one for the gravity. :smile:

I think you confused yourself by writing G instead of m in the acceleration equation, which made you think you'd already written the gravity equation! :confused:

Oh, and you've been asked for angular speed, not the ordinary speed, v.
 
  • #4
So you're saying the other equation I need is...

F = G[tex]\frac{m_{1}m_{2}}{r^{2}}[/tex]

So at this point I have...

F = 3.977 * 10[tex]^{30}[/tex][tex]\frac{v^{2}}{10000}[/tex]

and...

F = 6.67*10[tex]^{11}[/tex][tex]\frac{3.988*10^{30}*m_{2}}{r^{2}}[/tex]

Now what do I do. :/
 
  • #5
Well why use the mass of the planet for the centripetal force equation?

For the gravitational force you're looking at some mass, whatever it may be, that you denoted m2, located at the surface of the neutron star

This same mass is experiencing the centripetal force, which is being caused by gravity.
 
  • #6
the7joker7 said:
So you're saying the other equation I need is...

F = G[tex]\frac{m_{1}m_{2}}{r^{2}}[/tex]

Yes! :smile:

Your first F, [tex]m_2v^2/r[/tex], is the force needed to keep a mass [tex]m_2[/tex] in that circle of radius r.

Your second F is the gravitational force on a mass [tex]m_2[/tex] at a distance r.

You need the actual force to be equal to the force that's needed! :smile:

So you just make them equal.

(I suspect what you're finding confusing is that there seem to be two forces for no reason. It would be clearer if you went back to Newton's second law, force = mass x acceleration, or F = m x A, and wrote A = [tex]v^2/r[/tex]: same equation, but only one F, so less confusing.)
 
  • #7
Yeah, if you write the two equation symbolically, and keep your masses straight, it should become clear.
 

Related to Force required to keep objects in orbit.

What is the force required to keep objects in orbit?

The force required to keep objects in orbit is known as centripetal force. It is the force that acts towards the center of the orbit and keeps the object moving in a circular path.

What factors affect the force required to keep objects in orbit?

The force required to keep objects in orbit is affected by the mass of the object, the mass of the celestial body it is orbiting around, and the distance between the two objects. The greater the mass of the objects and the smaller the distance between them, the greater the force required to keep the object in orbit.

How is the force required to keep objects in orbit calculated?

The force required to keep objects in orbit can be calculated using Newton's law of universal gravitation: F = (G * m1 * m2) / r2, where F is the force, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them.

Can the force required to keep objects in orbit change?

Yes, the force required to keep objects in orbit can change if there are changes in the mass or distance of the objects. For example, if the mass of the object in orbit increases, the force required to keep it in orbit will also increase. Similarly, if the distance between the two objects increases, the force required to keep the object in orbit will decrease.

What happens if the force required to keep objects in orbit is not enough?

If the force required to keep objects in orbit is not enough, the object will not be able to maintain its orbit and will either fall back to the surface of the celestial body it is orbiting or move away from it. This is known as an unbound or unstable orbit.

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