Comet Physics

[Frida Space]

Hi everyone. I asked myself if an astronaut could escape the Sphere Of Influence of a comet just by jumping from its surface.

I took comet 67P/Churyumov-Gerasimenko as an example. The mass of the comet is approx. 3.14 x 10E12 kg. Its radius is 2000 m long (considering it a spherical object to make things easier).

The acceleration g on the comet is then equal to:

g = G x (m / rE2) x = 6.67428 x 10E-11 x 3.14 x 10E12 kg / 4000000 m = 5.24 x 10E-5

From this value we can calculate the escape velocity Ev of the comet.

Ev = √ (2Gm / r) = 0,4578 m/s

An average man on Earth jumps 0.2 m high. That means that:

√ 0.2 x 2 x 9.81 = 1.98 m/s

An average man jumps at an initial velocity of 1.98 m/s. That's almost 4 times the escape velocity, which means that one could easily escape the comet's SOI just by jumping. However, I tried to proof-check using other formulas, and something didn't quite work.

First, I calculated the SOI's radius:

Rsoi = a x ( m / M ) E2/5

where M is the mass of the Sun (1.99 x 10E30 kg) and a is the semi-major axis (518 120 000 000 m)

That makes a 37 408 m radius. This makes me think that, jumping from the surface, you would get an apoapsis far higher than 37 km, which would mean that you would be on an escape trajectory from the comet.

However, this is were the calculations say something different.

I simulated the astronaut's jump. We know his speed V at apoapsis will be 0 m/s, while his initial speed V0 will be 1.98 m/s. We also know that g on the comet is equal to 5.25 x 10E-5. Using this information, I easily calculated the height H reached.

H = V0 squared / 2 x g = 1.98E2 / 2g = 37 377 m

That is just inside the SOI radius, while I was expecting it to be far far beyond.

Can someone please explain me where I made a mistake? Thanks!!

Edited September 10, 2014 by Frida Space

[Jesrad]

You are making the assumption that gravity from the comet stays constant the further you get from it. This:

H = V0 squared / 2 x g = 1.98^2 / 2g = 37 377 m

is inapplicable in this context, because g varies with H.

[Frida Space]

You are making the assumption that gravity from the comet stays constant the further you get from it. This:

is inapplicable in this context, because g varies with H.

That's true. What a silly mistake. Thanks for the help!

[christok]

More problems with your analysis:

- An average man jumps way higher than 0.2m, which is only 20cm. You may have misread 20in. Try this table for better values: https://en.wikipedia.org/wiki/Vertical_jump#Vertical_jump_norms

- You need to account for solar tides, especially in borderline cases, so you will have a minimum and maximum value depending on where the body is located and where the person is standing.

Nitpicking:

- You should add a reasonable minimum mass for the spacesuit since it's significant relative to the mass of a man. (Including movement restrictions due to the suit would be difficult, so I recommend neglecting that on the basis that you may be using a more advanced form-fitting suit.)

- Energy makes mechanics easy. Use energy instead of acceleration. Your centre of mass is accelerated over the same distance on Earth as on the asteroid, which is the distance between where you start crouched and the point where your feet no longer touch the ground. Work is force times distance and the force your legs exert is the same in both cases, so work is the same. By the work-energy theorem, (sub-)orbital energy gained is the same. Increase in E_specific ~= 0.5m * g on Earth, so E_specific = 0.5 * 9.8 = 4.9 J/kg is gained on the asteroid as well. Add that to the potential energy between the surface and the centre of the asteroid to work out how high your new apoapsis is.

[Frida Space]

More problems with your analysis:

- An average man jumps way higher than 0.2m, which is only 20cm. You may have misread 20in. Try this table for better values: https://en.wikipedia.org/wiki/Vertical_jump#Vertical_jump_norms

- You need to account for solar tides, especially in borderline cases, so you will have a minimum and maximum value depending on where the body is located and where the person is standing.

......

thank you very much! My knowledge in physics is still very limited (I'm 16 years old) but I really want to improve it. Thanks again!

Edited September 10, 2014 by Frida Space

[christok]

Edit: Oops, I wrote megametres instead of kilometres. That makes the numbers significantly different but the conclusion is the same: Your astronaut escapes the surface with about 0.9J/kg specific energy over and above that required for escape velocity.

(Continuing the discussion here rather than in PM because it's relevant to everyone interested.)

I may well be wrong about the 20cm/20in confusion. If a NASA scientist said the centre of mass jumps up by 20cm on average and only the feet go up by much, you should use their value rather than some guy on the internet's.

We already know potential energy at the top of the jump in each case is equal to the kinetic energy at the bottom. However, there are two important caveats:

1) This is energy relative to the surface and not energy relative to the centre of the planet(esimal), which is what we'll be using for the orbit calculations.

2) g * m * h is an approximation which is only valid if the height is very small relative to the distance to the centre of the planet(esimal).

We can use g * m * h for the jump on Earth because 0.2m << 6Mm. Instead of regular kinetic and potential energy, we can use specific energy which is the energy divided by the mass and completely ignore the mass of the astronaut.

epsilon = specific energy = 0.5 * V^2 - mu / r, where mu = G * M.

So at the starting point, epsilon = KE - mu / 2km on the comet where KE = g_earth * h_earth is the specific kinetic energy at the start of the jump and equals potential energy relative to the surface for the Earth jump. Thus epsilon = 1.96 J/kg - mu_comet / 2km = 1.96J/kg - 3.14 * 10E12 kg * 6.673×10^-11 m^3 s^-2 kg^-1 / 2km = 1.96J/kg - 1.05J/kg unless I made a mistake somewhere.

Now we have a little shortcut. I did say energy makes it easy! You don't need to calculate the height you jump to. If (specific) orbital energy = 0, you have a parabolic trajectory. If it's < 0, you have an elliptical or circular orbit. If it's positive, the trajectory is hyperbolic and therefore we know that the astronaut escapes.

Edited September 12, 2014 by christok
Fixed the Mm/km slip-up.

[Frida Space]

So at the starting point, epsilon = KE - mu / 2Mm on the comet where KE = g_earth * h_earth is the specific kinetic energy at the start of the jump and equals potential energy relative to the surface for the Earth jump. Thus epsilon = 1.96 J/kg - mu_comet / 2Mm = 1.96J/kg - 3.14 * 10E12 kg * 6.673×10^-11 m^3 s^-2 kg^-1 / 2Mm = 1.96J/kg - 1.05 * 10^-3 J/kg unless I made a mistake somewhere.

Thank you, I understood everything but one thing: why did you put 2 Mm as the value of r? I get that 6 Mm is the radius of Earth, but 2 Mm isn't the radius of the comet. Can you explain me just that bit? Then I'll let you go :D (thanks again!!!)

[christok]

Thank you, I understood everything but one thing: why did you put 2 Mm as the value of r? I get that 6 Mm is the radius of Earth, but 2 Mm isn't the radius of the comet. Can you explain me just that bit? Then I'll let you go :D (thanks again!!!)

Check the note right at the top of the post. I accidentally said 2000km instead of 2000m because I was tired and when I noticed I put a note there rather than edit the whole thing because, again, I was tired. The value then becomes 1.96J/kg - 1.05J/kg, hence the 0.9J/kg specific energy.

[Frida Space]

Check the note right at the top of the post. I accidentally said 2000km instead of 2000m because I was tired and when I noticed I put a note there rather than edit the whole thing because, again, I was tired. The value then becomes 1.96J/kg - 1.05J/kg, hence the 0.9J/kg specific energy.

I get it, thanks again.