J
jim
Guest
What are the perihelion and aphelion speeds of Mercury? What are the perihelion and aphelion distances of this planet? Compute the product v * r at each of these two points and interpret your result physically.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
You should upgrade or use an alternative browser.
J
jondalar469
Guest
Mercury was at perihelion 27 January 2008 at 18h UT, or JD 2454493.25. The semimajor axis, a, is 0.38709927 AU, and the eccentricity, e, is 0.20563593.
The period of the orbit is found from
P = (365.256898326 days) a^(3/2) = 87.9695 days
The aphelion occurs half the period later, so the aphelion following the mentioned perihelion occurs on JD 2454537.23, which is 11 March 2008 at 17:38:02 UT.
The perihelion distance, rp, is found from:
rp = a (1 - e) = 0.38709927 (1 - 0.20563593) = 0.30749775 AU
The aphelion distance, ra, is found from
ra = a (1 + e) = 0.38709927 (1 + 0.20563593) = 0.46670079 AU
One AU is equal to 1.49597870691E+11 meters, so
a = 5.7909226E+10 meters
rp = 4.6001008E+10 meters
ra = 6.9817444E+10 meters
The Vis Viva equation says:
v = sqrt { GM [ 2/r - 1/a ] }
Where...
GM = 1.32712440018E+20 m^3 sec^-2
The perihelion speed is, then,
v(r = rp) = 58977 m/s
v(r = ra) = 38858 m/s
At perihelion, the angular momentum per unit mass will be
(v)(r) = (58977 m/s)(4.6001008E+10 meters) = 2.713E+15 m^2/sec.
At aphelion, the angular momentum per unit mass will be
(v)(r) = (38858 m/s)(6.9817444E+10 meters) = 2.713E+15 m^2/sec.
Hmm. The angular momentum is the same at both places. I wonder if that fact has any significance?
The period of the orbit is found from
P = (365.256898326 days) a^(3/2) = 87.9695 days
The aphelion occurs half the period later, so the aphelion following the mentioned perihelion occurs on JD 2454537.23, which is 11 March 2008 at 17:38:02 UT.
The perihelion distance, rp, is found from:
rp = a (1 - e) = 0.38709927 (1 - 0.20563593) = 0.30749775 AU
The aphelion distance, ra, is found from
ra = a (1 + e) = 0.38709927 (1 + 0.20563593) = 0.46670079 AU
One AU is equal to 1.49597870691E+11 meters, so
a = 5.7909226E+10 meters
rp = 4.6001008E+10 meters
ra = 6.9817444E+10 meters
The Vis Viva equation says:
v = sqrt { GM [ 2/r - 1/a ] }
Where...
GM = 1.32712440018E+20 m^3 sec^-2
The perihelion speed is, then,
v(r = rp) = 58977 m/s
v(r = ra) = 38858 m/s
At perihelion, the angular momentum per unit mass will be
(v)(r) = (58977 m/s)(4.6001008E+10 meters) = 2.713E+15 m^2/sec.
At aphelion, the angular momentum per unit mass will be
(v)(r) = (38858 m/s)(6.9817444E+10 meters) = 2.713E+15 m^2/sec.
Hmm. The angular momentum is the same at both places. I wonder if that fact has any significance?
E
Eddie
Guest
sounds like homework to me
D
Donald S
Guest
You got me there Bubba. I have a hardtime calculating my age (82, I thimk).