Gravitation (just to check if calculation is right)

AI Thread Summary
The discussion focuses on calculating the speed of two neutron stars as they collide, given their mass and initial separation. The initial calculation used the formula v = sqrt(2Gm2/R^2), but there was confusion regarding the correct formula, which should be v = sqrt(2Gm2/R). The total work done on the stars is derived from the gravitational force, leading to a kinetic energy equation for each star. The potential energy change is expressed as ΔU = -GmM(1/r_f - 1/r_i), which is crucial for understanding the stars' movement. The conversation highlights the importance of correctly applying gravitational equations in astrophysical calculations.
Hollysmoke
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I was wondering if someone could tell me if I did this right:

Two neutron stars are separated by a distance of 10^10m. They each have a mass of 10^30kg and a radius of 10^5m. They are initially at rest relative to each other. How fast are they moving when they collide?

R = 2r + d = 2(10^5)+10^10

v = sqrroot (2Gm2/R^2), sub in all the numbers and v = 1.2x10^5m/s
 
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Hollysmoke said:
I was wondering if someone could tell me if I did this right:

Two neutron stars are separated by a distance of 10^10m. They each have a mass of 10^30kg and a radius of 10^5m. They are initially at rest relative to each other. How fast are they moving when they collide?

R = 2r + d = 2(10^5)+10^10

v = sqrroot (2Gm2/R^2), sub in all the numbers and v = 1.2x10^5m/s

Where did you get "v= sqrroot(2Gm2/R2"? Shouldn't that be v= sqrroot(2Gm2/R)?
 
Hmm I think I rearranged the equation wrong then. Thanks.
 
I did Fg = Gm1m2/r^2 = 1/2mv^2
 
Hollysmoke said:
I was wondering if someone could tell me if I did this right:

Two neutron stars are separated by a distance of 10^10m. They each have a mass of 10^30kg and a radius of 10^5m. They are initially at rest relative to each other. How fast are they moving when they collide?

R = 2r + d = 2(10^5)+10^10

v = sqrroot (2Gm2/R^2), sub in all the numbers and v = 1.2x10^5m/s
The total work done is the integral of the force from initial r to final r:

\int_{r_i}^{r_f}Fdr = \int_{r_i}^{r_f}\frac{Gm^2}{r^2}dr = Gm^2\left(\frac{1}{r_i} - \frac{1}{r_f}\right)

This becomes the kinetic energy of both stars, each of which has half this energy:

KE_{star} = \frac{1}{2}mv^2 = \frac{1}{2}Gm^2\left(\frac{1}{r_i} - \frac{1}{r_f}\right)

v = \sqrt{Gm\left(\frac{1}{r_i} - \frac{1}{r_f}\right)}

AM
 
We haven't done integrals in this unit so I don't know how to solve that method =(
 
Hollysmoke said:
We haven't done integrals in this unit so I don't know how to solve that method =(
All you need to know is that the potential energy is U = -GmM/r. The change in potential energy between two positions is just \Delta U = -GmM(1/r_f - 1/r_i). This is the change in potential energy of the system. In this case, the system consists of two stars which move toward each other with equal speed.

AM
 

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