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

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SUMMARY

In circular orbits, the kinetic energy is accurately represented by the equation Ek = 1/2mv^2, while in elliptical orbits, the equation Ek = GMm/2r is derived from gravitational force considerations. The distinction arises because the radius (r) in the gravitational equation refers to the distance from the central body, while in the kinetic equation, it pertains to the radius of curvature of the motion. Total mechanical energy, which includes both kinetic and potential energy, is conserved in both circular and elliptical orbits, as they are governed by gravitational interactions.

PREREQUISITES
  • Understanding of gravitational force equations, specifically F=GMm/r^2
  • Familiarity with kinetic energy equations, including Ek = 1/2mv^2 and Ek = GMm/2r
  • Knowledge of circular and elliptical orbits in celestial mechanics
  • Concept of total mechanical energy conservation in orbital dynamics
NEXT STEPS
  • Study the derivation of kinetic energy equations in different orbital contexts
  • Explore the principles of energy conservation in gravitational systems
  • Learn about the eccentricity of orbits and its implications on energy calculations
  • Investigate the differences between circular and elliptical orbits in terms of force and motion
USEFUL FOR

Astronomy students, physics educators, and anyone studying orbital mechanics will benefit from this discussion, particularly those focusing on the application of kinetic energy equations in different types of orbits.

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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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.
 
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?
 
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.
 
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?
 
jakeginobi said:
In

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

A circle is an ellipse with eccentricity 0.
 
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.
 

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