# Compare centripetal acceleration of Mars and Earth

• shnigglefratz
In summary, the question posed is about the comparison of the centripetal accelerations of Mars and Earth in their orbits around the Sun. The relevant equation is a=v^2/r and the concept is that Mars has a larger radius and therefore a larger perimeter than Earth. Using Newton's law of universal gravitation and Newton's second law, it is determined that the only factor that matters in comparing the two accelerations is the radius of their orbits, and therefore Mars has a larger centripetal acceleration.

## 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.

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?)

rock.freak667 said:
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 = $\frac{G(m_{Earth})(m_{Sun})}{r^{2}}$?

shnigglefratz said:
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.

rock.freak667 said:
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 = $\frac{G*m_{Sun}}{r_{Mars' Orbit}^2}$

shnigglefratz said:
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 aearth/amars then GMsun will cancel out.

rock.freak667 said:
Right, so you can see that if you take aearth/amars then GMsun 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?

That should be correct.

rock.freak667 said:
That should be correct.

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

wrong if radius is larger acceleration is less

notsosmartman said:
wrong if radius is larger acceleration is less

## 1. What is centripetal acceleration?

Centripetal acceleration is the acceleration experienced by an object moving in a circular motion. It is always directed towards the center of the circle and its magnitude is equal to v^2/r, where v is the velocity of the object and r is the radius of the circle.

## 2. How does centripetal acceleration differ on Mars and Earth?

The centripetal acceleration on Mars and Earth differs due to their varying masses and radii. Mars has a smaller mass and radius compared to Earth, which results in a lower gravitational pull and thus, a lower centripetal acceleration.

## 3. What is the formula for calculating centripetal acceleration?

The formula for calculating centripetal acceleration is a = v^2/r, where a is the centripetal acceleration, v is the velocity of the object, and r is the radius of the circular path.

## 4. How does the centripetal acceleration on Mars and Earth affect objects?

The lower centripetal acceleration on Mars means that objects will experience less force keeping them in their circular path compared to Earth. This can result in objects having a longer orbital period or being able to escape the planet's gravitational pull more easily.

## 5. Can centripetal acceleration be manipulated on Mars or Earth?

Yes, the centripetal acceleration on both Mars and Earth can be manipulated by changing the velocity or radius of the circular motion. For example, increasing the velocity or decreasing the radius will result in a higher centripetal acceleration, and vice versa.