Calculating Escape Velocity of Mars from the Sun's gravitational field

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1. May 6, 2013

QuickSkope

1) The problem:
Mars orbits the Sun at a distance of 2.3 * 10^10 m with a speed of 24000 m/s. If it's speed were increased to 30000 m/s, at a time when no other planet was nearby, would mars leave the Solar System.

2) Fc=Fg
Orbit formulas and such

3) I'm assuming you have to find the escape velocity of Mars, and use it to see if it gets out of the field. But I'm not sure, any help would be awesome :)

Edit 1) Fc=Fg
V^2=GM/r

30000^2=(6.67 * 10^-11)(1.98*10^30)/r

R= 1.467 * 10^11

I'm not sure where the edge of the solar system is, and I doubt that's right. But worth a shot.

Last edited: May 6, 2013
2. May 6, 2013

SammyS

Staff Emeritus
Try using Kepler's third law .

3. May 6, 2013

QuickSkope

I have yet to learn Keplers law, so looking it up tells me:

The*square*of the*orbital period*of a planet isproportional*to the*cube*of the*semi-major axis*of its orbit.

I'm not really sure how I'd use that, but I'll find the current period of one rotation of Mars.

Period of Mars 6.0 * 10^7

Upon further investigation, we learned the formula (4(pi^2)r/T^2) = (GM/r^2), which is a representation of Kepler third law.

Also, is the method I used not a simplification of Keplers Third Law?

Last edited: May 6, 2013
4. May 6, 2013

SammyS

Staff Emeritus
You're right.

Using Kepler's 3rd Law was a silly idea. The 3000 m/s doesn't refer to a circular (or nearly circular) orbit.

The formula (4(pi^2)r/T^2) = (GM/r^2) that you mention is just from taking v2=GM/r and dividing by r . (FC = FG)

Do you know the escape speed for Mars orbit as regards escaping from the Sun's gravity, or know how to get it?

5. May 6, 2013

QuickSkope

I can do it on a projectile that is leaving a planet, not sure how I'd do it on two planets.

6. May 6, 2013

QuickSkope

The EP and Ek in the 2nd are 0, right?

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Last edited: May 6, 2013
7. May 6, 2013

QuickSkope

Last edited: May 6, 2013
8. May 6, 2013

SammyS

Staff Emeritus
How about from one planet with the mass of the Sun and radius equal to Mars's orbit?

9. May 6, 2013

QuickSkope

Not sure I quite follow that. Are my other calculations not correct?

I did math in the above and got escape velocity of 33888 m/s.

Also, the escape velocity to hit something from the surface of mars so that it does not come down is 2.4 km/s. If that helps at all.

Last edited: May 6, 2013
10. May 7, 2013

SammyS

Staff Emeritus
The 338888 m/s is what I was asking. (Actually, using the figures given in the problem, I got 33941 m/s.)

For an object that is at a distance of 2.3 * 1010 m from the Sun, 338888 m/s is the escape velocity for escaping the Sun. I.e. 338888 m/s is the velocity that Mars needs to escape the Sun's gravity.

11. May 7, 2013

QuickSkope

And so by that, if it is only travelling at 30000 m/s, it will in fact not escape the Suns gravitational field, and thus remain in the solar system?

12. May 7, 2013

SammyS

Staff Emeritus
Correct.

There are other ways to show it.

They agree.

13. May 7, 2013

QuickSkope

Awesome, thanks for the help :D