Calculating the Velocity and Distance of Two Colliding Planets using Gravity

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Homework Help Overview

The discussion revolves around calculating the velocity and distance of two colliding planets, represented by two spherical pieces of rock with given masses and radii, initially stationary and separated by a specified distance. The primary forces considered are gravitational, and the problem involves concepts from mechanics, particularly energy conservation and momentum.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between gravitational potential energy and kinetic energy, attempting to equate them to find the speed of one mass at collision. There are discussions about the conservation of momentum and the need to calculate changes in potential energy as the masses move closer together.

Discussion Status

The discussion is active, with participants providing hints and guidance on the conservation laws involved. There is an ongoing exploration of the implications of the center of mass in the system and how it relates to the distances traveled by each mass. Some participants express confusion about certain aspects, indicating a productive dialogue without a clear consensus yet.

Contextual Notes

Participants note the ambiguity in whether the initial separation distance is measured center to center or edge to edge, which affects the calculations. There is also mention of homework constraints that may limit the methods available for solving the problem.

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Homework Statement



Two spherical pieces of rock, of masses m1=1.0×1010kg and m2=2.0×1010kg and both of radius r=1500m (2.s.f.) are in deep space a distance of R=1000km (2.s.f.) apart. The only force between them is gravity and they are initially stationary. Find the speed of m1 at collision ? and how much distance m1 travels before colliding.

Homework Equations


F=Gm1m2/R^2
U=Gm1m2/r
E=0.5mv^2

The Attempt at a Solution


For the first part i tried to find the difference in potential then equate that to the kinetic energy but its was in vain
 
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Hashiramasenju said:
For the first part i tried to find the difference in potential then equate that to the kinetic energy but its was in vain
That's certainly part of the solution. Hint: Besides energy, what else is conserved?

Show what you did.
 
Doc Al said:
That's certainly part of the solution. Hint: Besides energy, what else is conserved?

Show what you did.
momentum is conserved so m1v1=-m2v2

But how do you calculate the potential energy?

Thanks for the reply
 
Hashiramasenju said:
momentum is conserved so m1v1=-m2v2
Right.

Hashiramasenju said:
But how do you calculate the potential energy?
You gave the formula yourself:
Hashiramasenju said:
U=Gm1m2/r
It's missing a minus sign, by the way.
 
Doc Al said:
Right.You gave the formula yourself:
​
It's missing a minus sign, by the way.
so is it Gm1m2/(2r+R)=0.5m1v1^2

where R is the distance between the surface of the two rocks
 
Hashiramasenju said:
so is it Gm1m2/(2r+R)=0.5m1v1^2

where R is the distance between the surface of the two rocks
You need the change in potential energy, as the rocks move from initial to final position. And that will equal the total kinetic energy of both rocks, not just m1.

You need to combine that with conservation of momentum to solve for the final speed of m1.
 
Doc Al said:
You need the change in potential energy, as the rocks move from initial to final position. And that will equal the total kinetic energy of both rocks, not just m1.

You need to combine that with conservation of momentum to solve for the final speed of m1.
Thats what is confusing me so is it
Gm1m2/(2r+R) -Gm1m2/(2r)=0..5m1v1^2+0.5m2v2^2
 
Hashiramasenju said:
Thats what is confusing me so is it
Gm1m2/(2r+R) -Gm1m2/(2r)=0..5m1v1^2+0.5m2v2^2
Almost. You have the sign wrong on the left-hand side. (What you have now is negative.)
 
Hashiramasenju said:
Thats what is confusing me so is it
Gm1m2/(2r+R) -Gm1m2/(2r)=0..5m1v1^2+0.5m2v2^2
Since we are not told whether the separation is center to center or edge to edge, it is not clear whether you need to account for the 2r in the starting separation. In any case, to two significant figures, 2r is negligible compared to R. So you may as well simplify the equation now and save yourself some work later.
 
  • #10
Doc Al said:
Almost. You have the sign wrong on the left-hand side. (What you have now is negative.)
OMG ! I got the answer. Thanks a lot.

So for the second part i got the answer by guessing that m1/m2=d2/d1
where d1 is the distance travveled by m1 and likewise for d2

and the answer was correct but i don't know why
 
  • #11
Hashiramasenju said:
OMG ! I got the answer. Thanks a lot.
Excellent.

Hashiramasenju said:
So for the second part i got the answer by guessing that m1/m2=d2/d1
where d1 is the distance travveled by m1 and likewise for d2

and the answer was correct but i don't know why
No guessing allowed! :-)

Hint: Where is the center of mass of this system? Does it change when they approach?
 
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  • #12
Doc Al said:
Excellent.No guessing allowed! :-)

Hint: Where is the center of mass of this system? Does it change when they approach?
I got it. But why doesn't the CM change ?
 
  • #13
Hashiramasenju said:
I got it. But why doesn't the CM change ?
If the two masses start at rest, the total momentum of the system is zero, yes? The total momentum of a composite system can (at least classically) also be computed as total mass multiplied by velocity of the center of mass.

If total mass is non-zero and momentum is zero, what does that tell you about the velocity of the center of mass?
 
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  • #14
jbriggs444 said:
If the two masses start at rest, the total momentum of the system is zero, yes? The total momentum of a composite system can (at least classically) also be computed as total mass multiplied by velocity of the center of mass.

If total mass is non-zero and momentum is zero, what does that tell you about the velocity of the center of mass?
Thanks !
 

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