In a shuttle craft of mass m = 2890 kg, Captain Janeway orbits a planet of mass
ID: 1353035 • Letter: I
Question
In a shuttle craft of mass m = 2890 kg, Captain Janeway orbits a planet of mass M = 8.59 1024 kg, in a circular orbit of radius r = 4.70 106 m.
(a) What is the period of the orbit? s
(b) What is the speed of the shuttle craft? m/s Janeway briefly fires a forward-pointing thruster, reducing her speed by 1.80%.
(c) Just then, what is the speed of the shuttle craft? m/s
(d) What is the kinetic energy of the shuttle craft at that moment? J
(e) What is the gravitational potential energy of the shuttle craft at that moment? J
(f) What is the mechanical energy of the shuttle craft at that moment? J
(g) What is the semimajor axis of the elliptical orbit now taken by the craft? m
(h) What is the difference between the period of the original circular orbit and that of the new elliptical orbit (Tnew - Told)? Note which orbit has the smaller period. s
Explanation / Answer
We first use the law of periods: T^2 = (4*pi^2/GM)r^3, where M is the mass of the planet and r is the radius of the orbit. After the orbit of the shuttle turns elliptical by firing the thrusters to reduce its speed, the semi-major axis is
a = -GMm/2E
where E = K + U is the mechanical energy of the shuttle
and its new period becomes T = (4*pi^2a^3/GM)^0.5
A.
T = [4*3.14^2*(4.7*10^6)^3/(6.67*10^-11*8.59*10^24)]^0.5 = 2673.29 sec
B.
v = 2*pi*r/T = 2*3.14*4.7*10^6/2673.29 = 11041.07 m/sec
C
v' = 0.982*v = 0.982*11041.07 = 10842.33 m/sec
D.
KE = 0.5*m*v'^2 = 0.5*2890*10842.33^2 = 1.698*10^11 J
E
Immediately after the firing, the potential energy is the same as it was before firing the thruster:
U = -GMm/r = -6.67*10^-11*2890*8.59*10^24/(4.7*10^6) = -3.52*10^11 J
F.
Adding these two results gives the total mechanical energy E = KE+UE
E = 1.698*10^11 -3.52*10^11 = -1.822*10^11 J
G.
a = -GMm/2E = -6.67*10^-11*8.59*10^24*2890/(-2*1.822*10^11) = 4.54*10^6 m
H.
Using Kepler’s law of periods for elliptical orbits (using a instead of r) we find the new period to be
T' = [[4*3.14^2*(4.54*10^6)^3/(6.67*10^-11*8.59*10^24)]^0.5] = 2537.95 sec
This is smaller than our result for part (a) by
T - T' = 2673.29 - 2537.95 = 135.34 sec
we see that elliptical orbit has a smaller period.
this problem was too long and If you have any other problem please post those in advanced physics section. Thanks.