Compare centripetal acceleration of Mars and Earth

[shnigglefratz]

Homework Statement

Is the centripetal acceleration of Mars in its orbit around the Sun larger or smaller than the centripetal acceleration of the Earth?

Homework Equations

I assume that the relevant equation is a=v$^{2}$/r since you are trying to compare the centripetal accelerations of Mars and Earth, however it is supposed to be a concept question according to the book so no data or variables should be needed to answer the question.

The Attempt at a Solution

So i know that Mars is farther away from the Sun than the Earth is, so r is obviously greater for Mars. Therefore the perimeter of Mars' orbit around the sun is greater than the perimeter of the Earth's orbit around the Sun. What I am confused about is how to figure out the velocity of Mars, or Earth for that matter. Is the velocity not needed to answer the question and I am merely missing something, or can you just assume that because Mars is farther away from the Sun it has a greater centripetal acceleration based on a trend?

If I did anything against the forum rules/policies please forgive me as this is my first post on Physics Forums.

 

[rock.freak667]

Considering the Earth only. The gravitational force between the sun and the Earth provides the centripetal force for the Earth. Can you find an expression for this acceleration?

(Hint: What does the law of universal gravitation say?)

 

[shnigglefratz]

Considering the Earth only. The gravitational force between the sun and the Earth provides the centripetal force for the Earth. Can you find an expression for this acceleration?

(Hint: What does the law of universal gravitation say?)

So with Newton's law of universal gravitation it's: F = $\frac{G(m_{Earth})(m_{Sun})}{r^{2}}$?

 

[rock.freak667]

So with Newton's law of universal gravitation it's: F = $\frac{G(m_{Earth})(m_{Sun})}{r^{2}}$?

Right, so the centripetal acceleration of the Earth is therefore? (Newton's second law)

Do the same with Mars now.

 

[shnigglefratz]

Right, so the centripetal acceleration of the Earth is therefore? (Newton's second law)

Do the same with Mars now.

so: m$_{earth}$a = $\frac{G*m_{Earth}*m_{Sun}}{r^2}$


divide by m$_{Earth}$: a = $\frac{G*m_{Sun}}{r_{Earth's Orbit}^2}$


Therefore for Mars the equation will be: a = $\frac{G*m_{Sun}}{r_{Mars' Orbit}^2}$

 

[rock.freak667]

so: m$_{earth}$a = $\frac{G*m_{Earth}*m_{Sun}}{r^2}$


divide by m$_{Earth}$: a = $\frac{G*m_{Sun}}{r_{Earth's Orbit}^2}$


Therefore for Mars the equation will be: a = $\frac{G*m_{Sun}}{r_{Mars' Orbit}^2}$

Right, so you can see that if you take a_earth/a_mars then GM_sun will cancel out.

 

[shnigglefratz]

Right, so you can see that if you take a_earth/a_mars then GM_sun will cancel out.

So the only factor that matters when comparing the two centripetal accelerations is the radius of their orbits. Therefore Mars has a larger centripetal acceleration because its radius is larger. Is that right?

 

[rock.freak667]

That should be correct.

 

[shnigglefratz]

That should be correct.

Thanks, I really appreciate all of the help . I just thought that the solution would involve more than one variable.

 

[notsosmartman]

wrong if radius is larger acceleration is less

 

[rock.freak667]

wrong if radius is larger acceleration is less

Wrong about what?