- #1
frisky
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How do you calculate the kinetic and potential energies of the Earth at the perihelion and aphelion?
Given that the Earth's orbit is nearly circular
GMm/R^2 = m(v^2)/R where M = sun's mass, m = Earth's R = orbital radius
so
v = √(GM/R)
The kinetic energy, E, would be
E = (1/2)m(v^2) = (1/2)GMm/R
The Earth's potential energy, U, would be
U = -GMm/R
However, the change in E between the perihelion and aphelion would be
(1/2)GMm/[(1/R) - (1/r)] where R and r are the respective orbital radii;
while the change in U would be
-GMm/[(1/R) - (1/r)] .
Given conservation of energy, the last two terms can not both be right.
This brings me back to the original question.
Given that the Earth's orbit is nearly circular
GMm/R^2 = m(v^2)/R where M = sun's mass, m = Earth's R = orbital radius
so
v = √(GM/R)
The kinetic energy, E, would be
E = (1/2)m(v^2) = (1/2)GMm/R
The Earth's potential energy, U, would be
U = -GMm/R
However, the change in E between the perihelion and aphelion would be
(1/2)GMm/[(1/R) - (1/r)] where R and r are the respective orbital radii;
while the change in U would be
-GMm/[(1/R) - (1/r)] .
Given conservation of energy, the last two terms can not both be right.
This brings me back to the original question.