1. (i) Using the knowledge that Mercury's semi-major axis is 5.79 x 1010 m and Mercury's orbit has an eccentricity of 0.206, calculate the perihelion and aphelion distances for Mercury. (ii) What are the perhelion and aphelion velocities of Mercury?
a = 5.79 x 1010 m
e = 0.206
rperi = a(1 - e) = 4.60 x 1010 m
rap = a(1 + e) = 6.98 x 1010 m
Using vis viva equation:
v2= G (M1 + M2) [2/r - 1/a]
where M1 is the sun's mass and M2 is mercury's mass (negligible).
v2peri = 6.67 x 10-11 x 1.99 x 1030 x [ 2./(4.60 x 1010) - 1./(5.79 x 1010)]
vperi = 59.0 km/s
v2ap = 6.67 x 10-11 x 1.99 x 1030 x [ 2./(6.98 x 1010) - 1./(5.79 x 1010)]
vap = 38.9 km/s
Note the large velocity difference between aphelion and perihelion for an elliptical orbit.
2. (i) Using the knowledge that Earth's semi-major axis is 1.496 x 1011 m and Earth's orbit has an eccentricity of 0.017, calculate the perihelion and aphelion distances for Earth. (ii) What are the perhelion and aphelion velocities of Earth?
a = 1.496 x 1011 m
e = 0.017
rperi = a(1 - e) = 1.47 x 1011 m
rap = a(1 + e) = 1.52 x 1011 m
Using vis viva equation:
v2= G (M1 + M2) [2/r - 1/a]
where M1 is the sun's mass and M2 is the earth's mass (negligible).
v2peri = 6.67 x 10-11 x 1.99 x 1030 x [ 2./(1.47 x 1011) - 1./(1.496 x 1011)]
vperi = 30.3 km/s
v2ap = 6.67 x 10-11 x 1.99 x 1030 x [ 2./(1.52 x 1011) - 1./(1.496 x 1011)]
vap = 29.3 km/s
Note the small velocity difference between aphelion and perihelion for a nearly circular orbit.
3. What is the semi-major axis of the least energy elliptical orbit of a space probe from Earth to Saturn? Relative to the Earth, what is the velocity of such a probe at the Earth's orbit? Earth's semi-major axis=1.496 x 1011 m; Saturn's semi-major axis = 1.427 x 1012 m.
a = 1/2 ( aearth + asaturn)
a = 0.5 x ( 1.496 x 1011 + 1.427 x 1012)
a = 7.883 x 1011 m
v2= G (M1 + M2) [2/r - 1/a]
where r = earth's orbit = 1.496 x 1011 m
v2 = 6.67 x 10-11 x 1.99 x 1030 x [ 2./(1.496 x 1011) - 1./(7.883 x 1011)]
v = 40.1 km/s
Earth's orbital velocity is approximately 30 km/s (see problem 2 above). So the relative velocity is 10 km/s in the direction of earth's orbit.
4. The Moon will move away from the Earth until it no longer lags the tidal bulges, and the angular momentum transfer will stop. Computer calculations indicate that this will occur at an Earth-Moon distance of 6.45 x 105 km. Calculate the Moon's orbital period then.
a = 6.45 x 105 km
M = 5.98 x 1024 kg
P2 = 4 pi2 a3 / (G M)
so, P = 5.15 x 106 s = 59.6 days