## A Problem of Gravity

Here’s a nice one for the cute problems folder.

Two spherically symmetric stars A and B of equal mass M and radius r have centres separated by a distance 6r. Ignoring any effects due to the orbital motion of the stars, determine a formula (in terms of G, M and r) for the minimum velocity with which material can be ejected from the surface of A so as to be captured by B.

### 60 Responses to “A Problem of Gravity”

1. andyinkuwait Says:

root(2GM/3r)

• Andy Says:

Ooops sorry can I change that to root(5GM/3r)

• telescoper Says:

Both wrong I’m afraid…

2. Andy Says:

…..aand again, due to more stupid arithmetic errors, changing to root(GM/r).

Sorry, must be all the student reports I have to write clouding my mind

3. Anton Garrett Says:

Sqrt(16GM/15r)

• Bryn Jones Says:

Yes, that’s what I get as well.

• Bryn Jones Says:

I’m wondering about the relevance of 7 points though, rather than \sqrt(16/15) points.

4. Toffeenose Says:

Yes, (16GM/15r)^1/2

5. Knud Says:

sqrt(4/3 GM/r)

6. Edd Says:

I really should check my working again after getting a different answer to everyone else – but sqrt(8/15 GM/r)?

• Edd Says:

Well I dropped a factor of 2 somewhere, so I should have been wrong with 16/15…

7. paulrho Says:

v=sqrt(16/15*GM/r)

• paulrho Says:

More exactly
v>sqrt(16/15*GM/r)

8. Will Says:

I have yet another answer, but I’ve been sloppy with signs, so I’m not going to post it. Intrigued to know what the answer is though.

9. telescoper Says:

Well, nobody has got it right yet!

The answer has obviously got to be some multiple of sqrt(GM/r) (by dimensions).

I don’t know how people got their various answers as nobody showed any working, but here is a hint:

Sketch a few equipotentials to help you identify where on the originating star the material must come from to satisfy the minimum condition. The answer to that is that it is the point on A’s surface that is furthest from the centre of B, i.e. A’s backside.

10. Phillip Helbig Says:

Is the ejection assumed to be perpendicular to the surface?

• telescoper Says:

Whatever direction minimizes the speed needed.

11. Miccou Says:

8/5

• telescoper Says:

Nope.

12. Anton Garrett Says:

Sqrt(16GM/15r) is the answer assuming that the particle is fired from the point on A which is closest to B, and is fired directly at B. This result is derived by calculating the gravitational potential due to both stars at this point on the surface of A, and also at the Lagrange point halfway between A and B; then equating the difference in potential at these two locations to the kinetic energy of a unit mass (ie the mass just makes it to the Lagrange point, with negligible velocity there, and the gravitational attraction of B then wins).

I did consider other start points on A, and other angles of firing, but reasoned informally that the test particle got the most “help” from B along the shortest path to it. Perhaps I should become more formal in future…

13. Bryn Jones Says:

sqrt(20GM/21r)

• telescoper Says:

We have a winner!

Although you should show your working!

• Bryn Jones Says:

The problem is unphysical, in the sense that the two stars will not be spherically symmetric because the potential of one will affect the potential of the other. My first attempt was wrong because, for simplicity, I considered an element of gas moving from the point on star A nearest to B.

• telescoper Says:

Yes, the shapes will actually be distorted but in the problem stated the key point is to realize the surface of the star A is not an equipotential…

• paulrho Says:

I tried similar – but had 22/21*… and it had to go through star A.
Something is not right???

• telescoper Says:

I am sure that Bryn’s answer is correct.

• Bryn Jones Says:

Show the working?

I’ll try to do that, but the equations may not display neatly.

Well, we can consider the problem from an energy viewpoint, and consider the gravitational potentials. An element of gas moving with the minimum possible velocity to cross the potential barrier would have the minimum kinetic energy needed to cross, and would do so from the point on the surface of star A where the gravitational potential is greatest (least deep). This will be on the surface of star A furthest from star B (because the contribution to the total gravitational potential from star B is least on A’s surface at that point). Call this point Q.

The element of gas would cross to star B through the point between stars A and B where the lowest maximum is encountered on the path: that means through the Langrange point which lies midway between the two stars (the star masses are equal). Call this point X.

The gravitational potential at point Q (the point on the surface of star A furthest from B) is

Phi_Q = (potential at Q due to star A) + (potential at Q due to star B)
= – GM/r – GM/(7r) = – 8GM/(7r)

The gravitational potential at point X (the midpoint) is

Phi_X = (potential at X due to star A) + (potential at X due to star B)
= – GM/3r – GM/3r = – 2GM/(3r)

The total mechanical energy of the element of gas is conserved in this case (ignore interactions between the element and other gas). So,

kinetic energy at Q + potential energy at Q = kinetic energy at X +
potential energy at X

kinetic energy at Q + potential energy at Q = kinetic energy at X +
potential energy at X

If the mass of the element of gas is m and its velocity is v_min when leaving star A at point Q,

(1/2) m v_min^2 + Phi_Q m = 0 + Phi_X m

So,

(1/2) v_min^2 = Phi_X – Phi_Q
= – 2GM/(3r) + 8GM/(7r)
= 10 GM/(21r)

This gives the minimum velocity to be

v_min = sqrt(20GM/21r)

• telescoper Says:

Excellent work. You win the star prize.

• Bryn Jones Says:

Unfortunately it appears to be an abnormally-shaped star prize.

• Anton Garrett Says:

The trajectory issue remains interesting.

• telescoper Says:

Yes, indeed. I hadn’t thought of that before.

• Bryn Jones Says:

Yes. The trajectory should be close to an ellipse close to star A, but would increasingly depart from the ellipse as the distance from A increases and the distance from B decreases. My strong inclination would be to investigate it numerically.

• Anton Garrett Says:

The thing is that the velocity at the Lagrange point is going to have a nonzero component perpendicular to the line joining the centres of the two ‘stars’. Yet Bryn’s calculation supposes that the velocity at the Lagrange point is epsilon…

• Bryn Jones Says:

Yes, that’s right. And as discussed in comments by ChrisC and Toffeenose below, the gas has to travel faster than the circular velocity at the surface of star A to avoid falling into star A.

• Anton Garrett Says:

Maybe the answer starts from the “far side ” of A, departs tangentially, but doesn’t go through the Lagrange point? (If it did then it would end up at the far side of B.)

• telescoper Says:

It does depart from the far side of the host star…..

14. ChrisC Says:

Clearly the difference in gravitational potenial is minimum between the Lagrange point and the surface on the far side of A. However, doesn’t that mean that the path to the mid-point lies below the surface of A? I haven’t actually worked out the path yet… In order to follow a trajectory to that Lagrange point, the optimum direction of launch will be horizontal, and to avoid a semi-minor (ish – it’s not elliptical) axis size of zero the test mass must arrive with a non-zero velocity. I therefore agree that the far side may lead to the lowest launch velocity but I think there must be an added term above the 20/21. I have work to do this afternoon so I will restrain myself from modelling it…

15. ChrisC Says:

Yes – if the material can fall through star A and out towards B, staying all the time on the line between centres, then 20/21 is the answer. A real trajectory needs a higher value. Textbook orbital dynamics only addresses the situation with one massive body, so simple considerations of periapsis/apoapsis can’t be used.

• telescoper Says:

That’s not what is envisaged. The material does not have to be directed towards B initially.

16. Toffeenose Says:

For a single spherical mass, v^2 must be equal to or greater than GM/r for the projectile not to fall back and hit the first spherical mass. For it to equal this minimum value, the initial velocity must be tangential to the surface of the first spherical mass. For a projectile starting on the extreme opposite side of the first spherical mass (relative to the second), v^2 must be even greater than this minimum value since the gravity of the second mass is increasing the probability for the mass to hit the first mass before it ‘clears’ the first mass. Thus, practically the factor cannot be as low as 20/21, and must be greater than 1…..

• Bryn Jones Says:

sqrt(GM/r) is the circular velocity for a particle in orbit about a mass M at a radius r. So, yes, the velocity would have to be larger than this to avoid hitting the star.

• telescoper Says:

Depends on the initial direction…

• Toffeenose Says:

For the projectile to leave from the ‘extreme far side’ of the sphere (Bryn’s point Q), no matter what the initial velocity angle is, the starting speed must be greater than sqrt(GM/r) for it not to hit the sphere…..(given the gravitational effect of the other sphere). As one moves away from Q, the situation becomes less clear…..but my intuition tells me that v>sqrt(GM/r) is valid for any positioning of Q. My incorrect specific estimate earlier suggests this also, since the estimate was too low, but still higher than sqrt(GM/r) – since it incorrectly assumed zero velocity at point X, and assuming a finite velocity at X would have made the estimate of the initial speed even higher!

17. Toffeenose Says:

Repeat the calculation of Bryan Jones above, but place Q at the point on sphere A where the tangent at Q passes through the centre of sphere B. This will ensure that a projectile from Q will not hit sphere A. Moreover, the minimum-velocity trajectory must pass through point X because of symmetry (as one must be able to fire a similar minimum-velocity projectile from sphere B to A). The only change in the calculation is that the potential at Q due to sphere B is now GM/r(35)^1/2, and not GM/r8. The final factor of 20/21 then becomes 2/(35)^1/2 plus 2/3, which is slightly bigger than 1.

• Toffeenose Says:

My recent estimate is in error, since it assumes that the kinetic energy at point X (midway between the spheres) is zero. This would not be the case, though I suspect that the minimum velocity trajectory would there have a zero velocity component along the axis joining the centre of the spheres…..

18. Miccou Says:

The surface of the star A must be solid and frictionless. The initial trajectory is tangential, if i’m correct.

• Miccou Says:

2 * [ (diif potential from A) + (diff potential from B) ]

2 * [ (1/r-1/3r) + (1/7r-1/3r) ] = 20/21r

• Miccou Says:

The end of the trajectory is at the Lagrange point. It is the same for the difference of potential relative to A, either you start from the far side or the near side. But the far side gives a boost to the initial B potential.

19. Colin Rosenthal Says:

I haven’t thought much about physics in recent years so I may be way off base, but surely these energy-arguments, while they can give bounds on the velocity, can’t actually answer the question, because that requires calculating an actual trajectory as a solution of the dynamical equations (and demonstrating that it passes from one star to the other without passing through the surface of the first star).

• telescoper Says:

Yes. Perhaps I’m missing something but it seems some commenters are assuming there’s only one trajectory with a given energy…

20. Chris Brunt Says:

I’m not convinced (yet) of Bryn’s passing-through-the-mid-point construction…

If you require that the gas leaves Star A on the far side and passes through the mid-point between the stars, surely that would pose a problem: the gas would have to pass through star A to get there if it went in a straight line. The stars are set up to have a finite size.

If you imagine instead that the mid-point is reached by (e.g.) departing the surface at some angle (e.g. tangentially), that would require some kind of “Figure 8” orbit to reach Star B, which seems impossible from an angular momentum viewpoint. Wouldn’t the angular momentum vector have to flip at the midpoint?

• Chris Brunt Says:

Perhaps if the gas travels in a corkscrew around the centre-line, it will behave…?

• Bryn Jones Says:

Yes, the solution sqrt(20GM/21r) is unphysical (as several people have argued above) because the gas would have to travel through star A. However, the gas does not need to travel on a straight line – the trajectory would be approximately elliptical in the vicinity of star A, with the departure from an ellipse increasing as the gas approaches the mid-point between the stars. The gas would then follow a similar, but mirror image, trajectory from the mid-point to star B. So the trajectory would be like half of a figure 8.

It is correct that angular momentum would not be conserved. If we consider angular motion about the centre of star A, the gravitational attraction of star B on an element of the gas will have a transverse component when the gas does not lie on the line connecting the centres of the the two stars.

• Chris Brunt Says:

Yet both forces from the stars are central. The Figure of 8 seems intuitively to be possible (e.g. considering time-reversal symmetry), but physics seems to say no. The gas would leave star A moving clockwise and arrive at star B moving anticlockwise.

For the corkscrew orbit, the orbital sense is not reversed (I think…) and it also seems intuitive that the restoring forces could lead to such an orbit. The combined stellar forces would (very crudely) attract the gas to a line joining the stars, with a secular evolution one way or another depending on which side of the mid-point you’re on. I think you’d have to fire it from star A with some velocity component towards star B… It’s not overly clear however that such an orbit would necessarily pass through the mid-point however…

• Bryn Jones Says:

(And by angular momentum not being conserved, I mean here the angular momentum of the gas that travels from star A to star B. The angular momentum of the entire system – gas plus star A plus star B – assuming it is an isolated system, will be conserved, of course.)

• Bryn Jones Says:

(Although for the minimum energy solution v = sqrt(20GM/1r) the gas will travel on a straight line, the line joining the centres of the stars. The discussion about other trajectories above refers to more physically realistic ones where the gas does not pass through star A, for v > sqrt(GM/r).)

• Chris Brunt Says:

Within the (somewhat idealised) limits of the problem as stated, surely it would be possible to load the gas particle with close-to-all of the angular momentum in the system?

• Chris Brunt Says:

Ah, apparently angular momentum is not conserved for the orbiting particle if the two masses are fixed in space (see Constants of Motion in Euler’s 3 body problem) so there’s no need to sweep it under the rug of total angular momentum. I didn’t think this through very carefully!

21. Toffeenose Says:

I doubt that there is an analytical solution, since it is a three-body problem – even though we are all assuming that the mass of the projectile is as close to zero as we wish! I suspect that the best we can say, without trial-and-error numerical trajectory analyses, is that the numerical factor in the solution must be greater than one for a physically realistic trajectory that does not intersect sphere A. We can also probably argue that any minimum-velocity solution must pass through X (the mid-point on the line between the centre of the spheres) at right angles to line joining the centre of the spheres…..

22. domenico Says:

It is an interesting problem.
I am thinking to a relate issue: it could be possible to evaluate the low energy transfer (for a spacecraft) between the Earth and a planet using the potential energy at the starting point (in the solar system near the Earth) and the potential energy at the final point (in a future time near the destination planet, for example some years in the future).
If the energy content of the fuel mass is known (and the energy of the final orbit is known) then with some tables of potential energy of the points in the solar system could be possible to obtain the minimal use of fuel along the interplanetary transport networks (or others trajectories) without evaluating pathways.