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Escape velocity from solar system

by Pepealej
Tags: earth, escape, solar, system, velocity
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Pepealej
#1
Jan7-13, 05:34 PM
P: 20
Hi. I've been wondering, how could I calculate the escape velocity for a body of mass m which is a certain distance r from the center of the earth to escape the entire solar system, assuming the distance from the center of mass (aprox. from the sun because it's about 99% of the mass of the solar system) is R.

I'm familiar with escape velocities and the whole gravitation basics, but I don't know how to grasp this. Could anyone help? Thanks :)
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Simon Bridge
#2
Jan7-13, 05:40 PM
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How would you normally do it?
Say you wanted the escape velocity from the Earth from a point below the surface of the Earth...
Pepealej
#3
Jan7-13, 05:49 PM
P: 20
From a point bellow? Hmmm, good question.

Gravity bellow the surface of the earth is proportional to r, not 1/r^2. Hence I should consider another expression for the potential energy.

But, anyway, this is not the case. The expression for the potential energy is unique. :S I'm getting confused.

Simon Bridge
#4
Jan7-13, 06:25 PM
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Escape velocity from solar system

Your residual speed at the surface needs to be the escape velocity from the surface.
But you can try using conservation of energy.

Gravity bellow the surface of the earth is proportional to r, not 1/r^2.
... assuming constant density for the Earth.

Basically you want to know what initial speed you have to be doing to escape a mass distribution - starting from within the distribution. This sort of thing is pretty hand-wavy since, for something like the solar-system, the specific path taken is important (i.e. if all the planets are on the other side of the Sun to the Earth when you start, you can start out slower maybe.)

So the concept of escape velocity is a bit vague in this scale.
Anyway, see:
http://adsabs.harvard.edu/abs/1987IAUS..117...39C

Using: hyperphysics solar system data
From the Earth/Moon orbit, you need 42.1kmps to escape the Sun's gravity.
If you used the entire mass of the solar system you'd get something like a 0.04% difference to this.

You need 11.2kmps to escape the surface of the Earth - but still bound to the Sun.
To escape the Sun as well, you need 42.1kmps residual speed ... so, <waves hands around> 53.3kmps would get you out of the solar system from anywhere bound to the Earth.

If you get too precise, you have to take the planetary dynamics into account.
Anything else is an approximation.
willem2
#5
Jan7-13, 08:19 PM
P: 1,403
Quote Quote by Simon Bridge View Post

You need 11.2kmps to escape the surface of the Earth - but still bound to the Sun.
To escape the Sun as well, you need 42.1kmps residual speed ... so, <waves hands around> 53.3kmps would get you out of the solar system from anywhere bound to the Earth.

If you get too precise, you have to take the planetary dynamics into account.
Anything else is an approximation.
There are two problems here.

1. The earth is already moving with about 30km/s around the sun. You only need to have about 12 km/s left after escaping the earth.

2. If you leave the surface of the earth with speed v (v > 11.2 km/s),
you will have kinetic energy (1/2)mv^2 with respect to the earth. The residual energy after getting far away from the earth will be (1/2) m (v^2 - 11.2^2), and this will need to be equal to the kinetic energy of the 12km/s residual speed you need,

so you get v^2 - 11.2^2 = 12^2 => v^2 = (11.2^2 + 12^2) => v = 16.4 km/s

This will be a speed of about 46.4 km/s with respect to the sun
Simon Bridge
#6
Jan7-13, 09:41 PM
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Well, I was hoping you'd be too distracted by the hand-waving ... you are right - the figures imply a shift between two non-inertial reference frames. The Earth is turning too ... so you'd start with the tangential velocity of the Earth's surface when escaping the Earth?
Pepealej
#7
Jan8-13, 04:47 PM
P: 20
Ok, now I srart to see things clearly :)

I've come up with the value of 42.1 kmps as to being the escape velocity of the solar system from the earth, but I've done so only considering there's a gravitational force: the sun's. I did:

1/2mv-GMsm/distance from earth to sun=0

But, if I'm sitting at the surface of the earth, won't the earth's gravitational force be significant as to taking it into account in the gravitational potential energy? If I do so I get a value of 43.6 kmps, and not 42.1 kmps.

It's quite disturbing since 42.1 kmps is the value I see in every webpage. Is my last reasoning right? Thanks :)

PS: This is without taking into account what willem2 said. But I think it's totally correct and I'll take it into account when I clear my previous doubts off :)
Simon Bridge
#8
Jan8-13, 06:19 PM
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If you were sitting at the surface of the Sun, and the Earth was in your direction of flight, then the Earth will have a different effect to if it is opposite your direction of flight. This is why I talked about the exact path.

When you pass the orbit of the Earth, the masses of the Earth, Venus, Mercury, and the Sun are all inside your Gaussian surface... if you want to approximate things that way.

But in either case you can do a bit of math to work out how big the difference can get whether you include the effect of the Earth or not. Try it and see.
cjl
#9
Jan8-13, 11:31 PM
P: 1,023
To calculate the escape velocity from the solar system, you can pretty much neglect everything except the sun (assuming you start far from any other objects). The cumulative mass of all the planets isn't really enough to worry about unless your trajectory takes you fairly close to them. If your trajectory does end up coming close to a planet, you can still solve everything just using the math for a two body problem - you basically just treat it as several sequential problems rather than one large one. While you are close to the planet, you can ignore everything except the planet, and while you are far from any planets, you ignore everything except the sun.
Simon Bridge
#10
Jan8-13, 11:58 PM
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Or, in the situation outlined in post#1, the Earth is the only nearby object.
Enthalpy
#11
Jan11-13, 12:27 PM
P: 661
Agree with answer #5.

To add my 2 cents worth:
- 42.1km/s is sqrt(2) times 29.8km/s (circular orbit)
- The effect of Earth's attraction IS the 11.2km/s escape speed
- The real cost for a launcher depends on if it launches eastward to take advantage of Earth's spin, how strong it accelerates at low altitude (but it must resist the aerodynamic forces, which also give small losses), what ascent curve it takes...
- So delta-V is often given starting from Low-Earth-Orbit instead, because from there they depend less on launch details. Though, mistakes are easily made, like inital speed and speeds squared...

Escaping the Solar system is perfectly accessible to our present launchers, but reasonable mission planners wouldn't do it without the powerful help of Jupiter, which reduces the necessary launcher capacity or increases the mass allowed to the spacecraft, by a huge amount. Jupiter is used every time when a spacecraft is sent to the outer planets, or to a Solar polar orbit... Some crafts pass by Venus and Earth before Jupiter.

(Most) other missions would be unaccessible to our launchers, like a transfer to outer planets within reasonable time and capture there to orbit the planet, if not using the gravitation and speed of the planets and giant moons. For instance, orbiting Pluto is unaccessible, only an accelerated flyby helped by Jupiter is possible with chemical rockets.

I join an xls spreadsheet that (...hopefully) computes accelerated transfers to the outer planets NOT using Jupiter, which is unreasonable. You choose the speed increment from Low-Earth-Orbit, it gives the speed vs ground, C3 and Sun, and delay to the planets and the arrival speed outside their gravitation potential. Don't expect more details from me, I made it for a different purpose.

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