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I was recently learning about gravitational fields in high school (senior year), when my physics teacher presented us with a problem that we were supposed to analyze the energy of the particle qualitatively. However it struck me on how to find the minimum velocity.

I asked my physics teacher, and he told me it used some calculus. He didn't explain further, but I am curious for an answer.

It goes something like this :

Two planets A and B are separated by a distance of R. A stone (point mass) is thrown from the surface of planet A towards planet B. Mass of A and B are known. Radius of planet A and B are the same and known. Find the minimum velocity to throw the stone from planet A to B. Mass of stone is also known.

This question may be incomplete in known variables, because I thought about the problem and made my up own question.

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closed as off-topic by ACuriousMind, sammy gerbil, Gert, CuriousOne, knzhou Aug 2 '16 at 20:40

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  • $\begingroup$ Rocks from Mars have often been found on Earth. You need the stone to achieve escape velocity, as I'm sure you know. softschools.com/formulas/physics/escape_velocity_formula/90 Once you achieve escape velocity, depending on the gravitional field of the other planet, it's possible. $\endgroup$ – user108787 Aug 1 '16 at 14:24
  • $\begingroup$ i see, so we just calculate it using escape velocity formula and the rock is set to go? I am looking at the derivation of escape velocity, why is it when Kinetic energy >= Potential energy a particle has enough energy to escape the planet. What is so special about having a positive value for the total energy? $\endgroup$ – Andy Robert Aug 1 '16 at 14:50
  • $\begingroup$ Basically, when you go up, you gain potential energy. You can only gain it by losing kinetic energy. Kinetic energy can never be negative, so you can't gain more in potential energy than you have available kinetic energy to lose. (Note that the conventional assignment of "gravitational potential energy at $\infty$ is $0$" is purely a convention. The actual value of potential energy isn't relevant in orbital dynamics: only the difference in potential energy between one place and another.) $\endgroup$ – Martin Kochanski Aug 1 '16 at 15:18
  • $\begingroup$ The rock is set to go, and because you are assuming the planets A and B are identical, I assumed that the gravitional field of B would pull it towards it. There are other factors, such as getting to where you want to go in a reasonable time. If you look up the flight path from Earth to Saturn of the Cassini space probe, you will see how it's speed was built up by swinging past Venus ,a few times I think and then Earth. $\endgroup$ – user108787 Aug 1 '16 at 15:21
  • $\begingroup$ @Obliv, then why did the question come up on the first posts review queue? $\endgroup$ – heather Aug 1 '16 at 16:05
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You've formulated the question very well, but you have left out just one variable. (Trying to understand something that is not in your grade level is a major crime, but that is your teacher's fault: he hasn't properly taught you not to think).

The way to split up the problem is:

  1. Get from the surface of planet A to "outer space" in the vicinity of planet A.

  2. Have enough velocity left over to get to the vicinity of planet B.

  3. Fall onto planet B.

It is convenient to consider "outer space" as being infinity: it makes the calculations easier and it makes only the tiniest difference to the figures.

1. From the surface of A to outer space

The initial velocity required to get from the surface of a planet to "infinity" is called the escape velocity. For the Earth I find it convenient to remember it as $11$km/sec, or as $\sqrt{2}$ times the orbital velocity of a very low orbit, which I remember as $8$km/sec. (That $\sqrt{2}$ is true for any gravitating body, anywhere). You can look up the exact figures in Wikipedia, but I hope you don't. Despite what your teacher thinks, science is not about blind obedience to established authority.

According to Newton's law of gravitation, the gravitational force of a planet of mass $M$ on an object of mass $m$ at a distance $r$ from its centre is $$-\frac{GMm}{r^2}\text{,}$$ where $G$ is the Newton's gravitational constant, which is the same always and everywhere. You can look it up, and also celebrate the fact that out of all the constants of nature it is the least accurately known.

The minus sign is because it makes most sense to treat all distances, velocities, accelerations and forces as acting upwards - and of course gravity pulls downwards.

Now, even at your grade you should know that the work performed by a force is equal to the force times the distance moved. So on a tiny bit of the object's journey up from the planet's surface (a distance $\Delta{r}$, say) the work performed by gravity is $-\frac{GMm}{r^2}\Delta{r}$. Adding up all the little pieces, the total amount of work performed by gravity on the object's journey from the surface into outer space is $$-\int_{r_A}^{\infty}\frac{GMm}{r^2}d{r}\text.$$ (If your teacher says that integration is beyond your grade level, strangle him. Integration is easy. Get an elementary calculus book and read it for fun and see).

Doing the integration, the total work done by gravity turns out to be $-\frac{GMm}{r_A}$.

If you launched your particle with a velocity $v$, that means that it started with a kinetic energy of $\frac12{m}v^2$. When it gets to outer space, the work done by gravity means that the resultant kinetic energy is $$\frac12{m}v^2-\frac{GMm}{r_A}$$

and you'll see that this makes sense, because for small $v$ it's negative (the particle never gets that far), for $v$ equal to the escape velocity it's exactly zero (the particle escapes but that's that), and for larger $v$ it's positive, so there's still some kinetic energy left.

A couple of points:

  • There is a factor of $m$ in both halves of the equation. This shows that the mass of the particle isn't relevant to the dynamics of its motion.

  • If planet A is the Earth, you don't know $M$ without looking it up in a book, and you don't know the value of $G$ without looking it up in a book. That would be immoral. On the other hand, you could measure the radius of the Earth if you wanted (Eratosthenes seems to have been the first to do this, and it's quite a doable experiment for everybody), and you could also measure the acceleration due to gravity at the Earth's surface. You would therefore be able to use "acceleration = $GM/r^2$" to work out the value of the product $GM$, and thus be able to work out the escape velocity without looking up anything at all.

2. From outer space near planet A to outer space near planet B

I'll be much briefer here. Planet A and planet B are both (I hope) orbiting the Sun. If planet B is further away than planet A, you will need some extra kinetic energy to climb out of the Sun's "gravity well". If you prefer, you can think of it as needing "surplus velocity" after escaping from planet A.

I will now cheat and say that if you are going out from the Earth to Mars, you need to have $2.9$km/sec of velocity left over, once you have got to outer space, to get out from the vicinity of the Earth to the vicinity of Mars. You could do this working out for yourself, by deducing the acceleration due to the Sun's gravity at the Earth's distance from the Sun (using the length of the year) and comparing it to that at Mars's distance from the Sun (using the Martian year). But I do need to let you do some of the work yourself!

Just one other point: it isn't $11.2+2.9=14.1$km/sec you'd need to get to Mars. You need a starting kinetic energy which gets you to $2.9$km/sec when you get to "outer space", and because kinetic energy is proportional to the square of velocity, this means that you only need $11.6$km/sec to start with.

On the other hand, if Planet B is nearer to the Sun than planet A (Venus, for example), then you don't need any extra velocity at all. The escape velocity is enough.

The relative orbits of planets A and B are the variable that you left out of your question.

3. From outer space near Planet B to the surface of Planet B.

No extra velocity needed. Start at zero, and Planet B's gravity will carry you in all the way.


I've taken a long time over this because you sound like the sort of person who doesn't just want canned answers from books. Working things out for yourself is what science ought to be about (life, too). It's just unfortunate that so many schools seem to teach the opposite.

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    $\begingroup$ Excellent, I think Einstein said his education almost ruined his curiosity, or words to that effect. $\endgroup$ – user108787 Aug 1 '16 at 15:28
  • $\begingroup$ Thank you so much Martin, I'll try my best to understand. My teacher mentioned using integration and such. Is it all for the integration part? We just use integration to find the gravitational potential energy? $\endgroup$ – Andy Robert Aug 1 '16 at 15:37
  • $\begingroup$ An excellent answer! But I think you might be wrong when you say that you don't need any extra velocity if you are targeting Venus. A stone ejected from the Earth at escape velocity will go into orbit around the Sun, and there is no a priori reason to expect that the resulting orbit can get close enough to the Sun to fall into Venus' gravity well. To see this, imagine that the Earth is very small; then the escape velocity will be small too, and the resulting orbit will not deviate very far from the Earth's orbit. $\endgroup$ – TonyK Aug 1 '16 at 19:41
  • $\begingroup$ @Andy Robert: Yes, you only need integration for the gravitational potential energy. Once you have justified the formula to yourself you can try memorising it. Myself, I tend to forget the formula and need to work it out again by integration. $\endgroup$ – Martin Kochanski Aug 2 '16 at 10:14
  • $\begingroup$ @TonyK: Yes, that is a good point. One needs to have some velocity past escape velocity in order to lose enough orbital kinetic energy (energy of the orbit around the Sun) for the orbit to become an ellipse whose apogee is near the Earth and whose perigee is near Venus. The task would be interesting but a little involved to squeeze into a comment. Hohmann transfer orbits are the thing to read about here. I think my Mars answer may not have been quite right, for the same reason. $\endgroup$ – Martin Kochanski Aug 2 '16 at 10:19
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I will assume we are in an empty universe with only two spherical planets, planet A with mass $M_A$ and planet B with mass $M_B$. Both planets have a distance of $D$. Let us assume that you are standing exactly on the line connecting both planets' centres on the surface of planet A, which has a radius $x_{start}$. You are throwing the stone (mass $m$) with a velocity $v$ in the direction of planet B. Your stone now has an energy of: $$ E_{total} =E_{kin} + V(x_{start}) = \frac{1}{2} m v^2 - \frac{mG}{D} \left( \frac{M_A}{x_{start}/D} + \frac{M_B}{1-x_{start}/D} \right) $$ The first term is the kinetic energy, the second the potential energy within the gravitational field of both planets. The gravitational field of an object with mass $M$ is given by $$ - \frac{G M}{r} $$ where $r$ is the distance to the object.
Now, to escape the planet, your stone needs to work against the gravitational pull of planet A, so it will lose kinetic energy. (Energy is conserved, so the total energy will stay constant with potential and kinetic energy being converted into each other.) It will lose kinetic energy as long as the potential energy is growing. We can calculate where the kinetic energy is at its maximum (by requiring that the first derivative is zero): $$ x_E = \frac{D}{1+\sqrt{\frac{M_B}{M_A}}}$$ The potential energy at this point is: $$V(x_E) = - \frac{mG}{D} \left(M_A + 2 \sqrt{M_A M_B} + M_B \right)$$ Now the stone still needs to have kinetic energy left at this point (so that it is still flying towards planet B), thus we require: $$ E_{total} > V(x_E) \Leftrightarrow E_{kin} > V(x_E)- V(x_{start})$$ this gives the condition $$v^2 > - 2 \frac{G}{D} \left(M_A + 2 \sqrt{M_A M_B} + M_B - \frac{M_A}{x_{start}/D} - \frac{M_B}{1-x_{start}/D} \right)$$ Thus we can calculate for given values of the planet masses and distances what start velocity the stone needs to have.

This whole can be understood graphically by drawing the potential e.g. for $D=3$ and $5 M_A = M_B$:
Plot of potential
If we start e.g. on the left, the stone needs to climb up that potential hill (loosing its velocity in the direction of planet $B$ in the progress until it has gone over the maximum point of that hill).

Note that if the stone is not thrown along the connecting line between both planet centers, the problem becomes 2 dimensional and thus harder. Also, the presence of other objects in the surrounding space will make the calculation harder, but the principle is the same - the stone will only be able to make it to the other planet if its initial energy is higher than the highest potential energy it encounters on its journey. I will add a plot of the potential energy between the two planets soon.

I am happy about any corrections, additions or comments :)

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