Consider a spacecraft with \\empty\" mass Mf = 100: metric tons (one metric ton
ID: 1424048 • Letter: C
Question
Consider a spacecraft with empty" mass Mf = 100: metric tons (one metric ton equals 1000 kg, exactly), in the same orbit about the Sun as the Earth. That is, it circles the Sun at radius at = 1.495*10^8 km with orbital period 1.000 year = 3.156*10^7 s. To escape the solar system, the spacecraft must be boosted to 2^0.5 times its orbital speed, tangential to its orbit. This can be done with rocket engines with nozzle speed U = 4.000 km/s. Using these data, calculate the total fuel/oxidizer mass which must be expelled as exhaust to accelerate the spacecraft to the desired escape speed. Be careful with units here!
Explanation / Answer
Hi,
This problem can be solve through the principle of conservation of linear momentum. To do that we imagine the spacecraft as a system, which is composed by the fuel and the empty mass of the spacraft.
At the beginning both masses are moving together with a common speed (v), which is the orbital speed. This value can be obtained using the equations of circular movement. Of course this is only possible if we assume that Earth's orbit around the Sun (the one they say its the same of the spacecraft) is a circumference (1).
One of the equations of circular movement is:
T = 2R/v ; where T is the period of the movement, R is the radius of the circumference and v is the tangential speed.
In this case, we have the following:
v = 2R/T = 2*(1.495*1011 m)/(3.156*107 s) = 2.98*104 m/s
Now, at the end of the movement we have that the spacecraft travels at a higher speed (u), but it has expelled certain amount of mass (the fuel) which travels at its own speed (w).
As this two points in time are referred to the same system, they have the same amount of linear momentum. Therefore we have the following:
(M + dm)*v = M*(v+dv) + dm*(v-w) ; where M is the entire mass of the spacecraft except for a little fuel, dm is the mass of the fuel ready to be expelled in an instant of time, v is the orbital speed, dv is an instantaneus increase of speed and w is the speed of the fuel (which is negative because it points in the opposite sense of the movement of the spacecraft).
Once we work a little on the previous expression, we have the following:
Mdv = wdm :::: which is a differential equation that we can solve easily. One thing to have in mind is that dm can be expressed in terms of dM like this:
dm = -dM ; and the negative sign is due to the fact that the more fuel is expelled by the spacecraft the lower is its total mass.
If we integrate from the orbital speed (v) to the escape speed (u) and from the initial mass (empty mass + fuel mass) to the final mass (which I assume is the empty mass) we have the following:
dv = -w dM/M :::::::: u - v = w Ln[Mi / Mf] ; in this case: Mi = m0 + mf and Mf = m0
where m0 is the empty mass of the spacecraft and mf is the mass of the fuel expelled (2).
In the previous equation we have every value except for the escape speed, but this value can be calculated thanks to the relation given by the problem:
u = (2)^1/2 v = (1.4142)*(2.98*104 m/s) = 4.21*104 m/s
Finally, using the expresion from the integral we have the following:
u - v = w Ln[(m0 + mf)/ m0] :::::::::: mf = m0 [exp((u - v)/w) - 1]
The escape speed is w = 4 km/s = 4000 m/s = 4*103 m/s
mf = (100000 kg)*[ exp((4.21*104 - 2.98*104)/(4*103)) - 1] = (100000 kg)*(20.65) = 2065000 kg = 2065 metric tons.
NOTES:
(1) According to one of Kepler's Laws, Earth's orbit is an ellipse with low excentricity.
(2) This is only true if we assume that all the fuel is used for the spacecraft to escape the solar system.
I hope it helps