Physics Astronomy!

1. The same equation for circular orbit holds for elliptical orbit if the orbital radius in circular orbit is replaced by the semi-major axis in elliptical orbit.

Hence, GmM/a^2 = ma(2.pi/T)^2
where m and M are the masses of Ganymede and Jupiter respectively
G is the Universal Gravitational constant (= 6.67 x10^-11 N.m^2/kg^2)
a is the semi-major axis of the orbit
T is the period of Ganymede

T^2 = (4.pi^2/GM).a^3
i.e. T^2 = [4.pi^2/(6.67x10^-11 x 1.9x10^27)].(1.07x10^9)^3 s^2 = 3.816 x 10^11 s^2
T = 6.178 x 10^5 s = 172 hours

2. (a) Distance of spacecraft from centre of Mars
= (400 + 3390) km = 3790 km = 3.79 x 10^6 m
Acceleration = (6.67x10^-11).(6.42 x 10^23)/(3.79x10^6)^2 m/s^2
= 2.981 m/s^2

(b) (i) Potential energy = -(6.67x10^-11).(6.42x10^23)/(3.79x10^6) J
= -1.13 x 10^7 J
(ii) Total mechanical energy = -(1.13 x 10^7)/2 J = -5.65 x 10^6 J
(iii) Kinetic energy = (-5.65x10^6 - (-1.13x10^7)) J = 5.65 x 10^6 J

3. When the spacecraft is closest to Mars, its angular momentum
= (6.42 x 10^23) x (4538) x (1.71x10^5 + 3.39x10^6) kg.m^2/s
= 1.037 x 10^34 kg.m^2/s
When the spacecraft is farest from Mars, angular momentum
= (6.42 x 10^23)v(1.78x10^7 + 3.39x10^6)
= (1.36 x 10^31)v
where v is the speed of the spacecraft

By conservation of angular momentum,
1.037x10^34 = (1.36x10^31)v
v = 763 m/s

Note: You probably made a typing mistake on the mass of Mars, which should be 6.42 x 10^23 kg (see the same figure in Q2).

Actually 天同 is a professional physicist.

我也在等著睇嘢！~~~

這次或許要靠天同知識長幫幫忙～

2013-10-20 00:47:01 補充：
嗯~
佢真係識astronomy的野~
可能佢 undergraduate 果陣係讀 Physics~

2013-10-20 01:17:17 補充：
好野～
解答左啦～

^__^

2013-10-20 16:43:00 補充：
I see, he is a scientist. Nice to meet him~ ☆ヾ(◕‿◕)ノ

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