# Centripetal force and orbital motion...

Hi,

The force of gravity is the centripetal force when an object is in motion. But in projectile motion, where the speed is not enough to keep it in orbit, could the force of gravity be equated to centripetal force?

I'm trying to understand if all types of curved motion can be explained by centripetal force, or put differently, instances where the concept of centripetal force applies. For example, if a projectile is allowed to fall right through earth matter (instead of colliding with the ground (in our imagination, of course)), would it be able to orbit it in some kind of weird curve? Would the centripetal force apply here? If it can, then it must have to be different than the force of gravity, which is strange because centripetal force is always just a manifestation of some other force and the only force involved in projectile/satellite motion is gravity..

I hope you guys are getting my point..

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PhanthomJay
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Centripetal forces exist on any curved path, not just a circular path. For parabolic motion under gravity only, the centripetal force at any point is the component of the gravity force perpendicular to the tangent of the path at that point

• ViolentCorpse
For parabolic motion under gravity only, the centripetal force at any point is the component of the gravity force perpendicular to the tangent of the path at that point
Interesting!

There is, however, a radius term in the formula of centripetal force. What could be the radius of a parabola?

Chandra Prayaga
Interesting!

There is, however, a radius term in the formula of centripetal force. What could be the radius of a parabola?
In motion along any plane curve, there are two components of acceleration, tangential and normal. The tangential component is equal to dv/dt, where v is the speed. The normal component is equal to v^2/r where v is the speed and r is the radius of curvature. This is correct for any curve, parabolic, circular, or arbitrary.

• ViolentCorpse
In motion along any plane curve, there are two components of acceleration, tangential and normal. The tangential component is equal to dv/dt, where v is the speed. The normal component is equal to v^2/r where v is the speed and r is the radius of curvature. This is correct for any curve, parabolic, circular, or arbitrary.
Makes perfect sense! Thanks so much! :)

There's just one more thing I must ask: Is the centripetal force less than the force due to gravity when the curve of a projectile is parabolic, equal to gravity when it is circular and greater than gravity when it is elliptic?

Chandra Prayaga
Makes perfect sense! Thanks so much! :)

There's just one more thing I must ask: Is the centripetal force less than the force due to gravity when the curve of a projectile is parabolic, equal to gravity when it is circular and greater than gravity when it is elliptic?
1. Projectile motion with only gravity: The two components, tangential and normal, must together add as vectors to give the downward acceleration due to gravity, so the normal (centripetal) component will be numerically less than g.
2. Circular or elliptical motion: If we are still talking about motion on the earth, I don't see gravity alone being responsible for either circular or elliptical motion, so the comparison cannot be made.
3. Planetary motion with a single center of force: If the motion is circular, then it is also with constant speed, so there is no tangential component of acceleration, and the normal component is exactly equal to gravity. If the motion is elliptical, the gravitational force contributes to the normal component and the tangential component, so the normal component will be numerically smaller than the gravitational force/mass.

• ViolentCorpse
Thank you so much! :)