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

• QuickSkope
In summary, Kepler's Third Law can be used to find the escape velocity of a planet if it is in a certain orbit. Additionally, the method used to calculate the escape velocity is not an exact simplification of Kepler's Third Law.
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.

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QuickSkope said:
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.
Try using Kepler's third law .

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?

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QuickSkope said:
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?
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?

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

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

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QuickSkope said:
I can do it on a projectile that is leaving a planet, not sure how I'd do it on two planets.
How about from one planet with the mass of the Sun and radius equal to Mars's orbit?

SammyS said:
How about from one planet with the mass of the Sun and radius equal to Mars's orbit?

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.

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QuickSkope said:
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.
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.

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

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

There are other ways to show it.

They agree.

Awesome, thanks for the help :D

## 1. How do you calculate the escape velocity of Mars from the Sun's gravitational field?

The escape velocity of Mars from the Sun's gravitational field can be calculated using the formula: Ve = sqrt((2GM)/r) where Ve is the escape velocity, G is the gravitational constant, M is the mass of the Sun, and r is the distance between Mars and the Sun.

## 2. What is the gravitational constant and how is it used in the calculation?

The gravitational constant (G) is a fundamental physical constant that represents the strength of the gravitational force between two objects. It is used in the calculation of escape velocity because it relates the mass and distance of the objects to the strength of the gravitational force.

## 3. What is the mass of the Sun and how is it determined?

The mass of the Sun is approximately 1.989 x 10^30 kilograms. This value is determined through various methods such as studying the motion of other celestial bodies around the Sun and measuring the gravitational pull of the Sun on these objects.

## 4. How does the distance between Mars and the Sun affect the escape velocity?

The distance between Mars and the Sun (r) has a direct impact on the escape velocity. As the distance increases, the escape velocity decreases. This is because the gravitational force decreases with distance, making it easier for an object to escape the Sun's gravitational field.

## 5. Is the escape velocity of Mars from the Sun's gravitational field affected by other factors?

Yes, the escape velocity of Mars from the Sun's gravitational field can be affected by other factors such as the mass of Mars and the gravitational pull of other celestial bodies. However, these factors have a smaller impact compared to the distance between Mars and the Sun.

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