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tnutty
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Homework Statement
Mercury's orbital speed varies from 38.8km at aphelion to 59.0km at perihelion.
If the planet is 6.60*10^10m from the Sun's center at aphelion, how far is it at perihelion?
Homework Equations
v = sqrt(GM/r) (speed,circular orbit)
The Attempt at a Solution
V_a = velocity of aphelion = 38.8km = 3.88 * 10^4m
V_p = velocity of perihelion = 59.0km = 5.9 * 10^4m
V_a = sqrt(GM/r)
3.88 * 10^4m = sqrt(GM/r_a)
and solving for M gives me (I think)
M = (3.88 * 10^4m*r_a)^2 / G
V_b = sqrt(GM/r_p) ; where W_b = 5.9 * 10^4m
solving for r_p gives :
r_p = (3.88 * 10^4m *r_a ) / 5.9 * 10^4m
; where r_a = 6.60*10^10m;
and
r_p = 4.60 * 10^10
is this right?