With a circular orbit, when do I use 1/2mv^2

However, since the total mechanical energy (sum of kinetic and potential energy) is constant, energy can be exchanged between potential and kinetic energy, as long as the total remains constant. In a circular orbit, the kinetic energy and potential energy are equal, while in an elliptical orbit, they are not. Therefore, the conservation of energy applies to both circular and elliptical orbits, but the distribution of energy between kinetic and potential energy varies.
  • #1
jakeginobi

Homework Statement


At a circular orbit and at an elliptical orbit when do I use 1/2mv^2 instead of the kinetic equation from which I derived from F=GMm/r^2 which is Ek = GMm/2r

Homework Equations


F=GMm/r^2, Ek = GMm/2r,
Ek = 1/2mv^2

The Attempt at a Solution


For instance, when I tried to use the other kinetic energy equation (GMm/2r) to solve the kinetic energy of a satellite at perigee, it gave a totally different answer if I use 1/2mv^2
 
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  • #2
jakeginobi said:

Homework Statement


At a circular orbit and at an elliptical orbit when do I use 1/2mv^2 instead of the kinetic equation from which I derived from F=GMm/r^2 which is Ek = GMm/2r

Homework Equations


F=GMm/r^2, Ek = GMm/2r,
Ek = 1/2mv^2

The Attempt at a Solution


For instance, when I tried to use the other kinetic energy equation (GMm/2r) to solve the kinetic energy of a satellite at perigee, it gave a totally different answer if I use 1/2mv^2

You need to quote the precise question and show your working.
 
  • #3
Perhaps you're confusing the total mechanical energy which includes both kinetic and potential energy and is a constant over the whole orbit, with the kinetic energy alone at a particular location?
 
  • #4
Your kinetic equation is valid only for circular orbits. It is derived by balancing the forces.

GMm/r2 = m v2 / r

The problem is that the r on the left is the distance from the sun (or whatever) and the r on the right is the radius of curvature of the motion. These are only the same if the orbit is circular.
 
  • #5
In
Cutter Ketch said:
Your kinetic equation is valid only for circular orbits. It is derived by balancing the forces.

GMm/r2 = m v2 / r

The problem is that the r on the left is the distance from the sun (or whatever) and the r on the right is the radius of curvature of the motion. These are only the same if the orbit is circular.
Is total energy conserved in a circular orbit or just an elliptical orbit?
 
  • #6
jakeginobi said:
In

Is total energy conserved in a circular orbit or just an elliptical orbit?

A circle is an ellipse with eccentricity 0.
 
  • #7
jakeginobi said:
In

Is total energy conserved in a circular orbit or just an elliptical orbit?

In this system you have only gravitational potential energy and kinetic energy, so between those two quantities energy must be conserved for any orbit.
 

Related to With a circular orbit, when do I use 1/2mv^2

1. What is the significance of a circular orbit?

A circular orbit is an orbit in which an object travels around another object in a circular path. This type of orbit is important because it allows objects to maintain a relatively constant distance from the object they are orbiting, making it easier to calculate and predict their movements.

2. When do I use the equation 1/2mv^2 for a circular orbit?

The equation 1/2mv^2, also known as the kinetic energy equation, is used to calculate the amount of energy an object has due to its motion. In a circular orbit, this equation is used to calculate the kinetic energy of the object as it moves around the central object.

3. Can the equation 1/2mv^2 be used for any type of orbit?

Yes, the equation 1/2mv^2 can be used for any type of orbit, as long as the object is in motion. It is commonly used in circular orbits because the velocity is constant, making it easier to calculate the kinetic energy. However, it can also be used for elliptical or parabolic orbits.

4. How does the mass of the object affect the use of the equation 1/2mv^2?

The mass of the object does not affect the use of the equation 1/2mv^2. This equation is used to calculate the kinetic energy of an object, which is dependent on its velocity and not its mass. However, the mass of the object will affect the overall energy of the orbit due to the gravitational pull of the central object.

5. Is the equation 1/2mv^2 the only equation used for circular orbits?

No, the equation 1/2mv^2 is not the only equation used for circular orbits. Other equations, such as Newton's law of universal gravitation and the centripetal force equation, are also important for understanding and calculating circular orbits.

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