How Fast Can a Satellite Traverse 1500 km Using Limited Commands?

[kosherkittens]

No source of friction in space to slow you down. So if you want to stop motion, you have to rely on other forces. Consider the problem of a moving satellite using only the following commands:

1) turn on the main thruster

2) tun off the main thruster

3) rotate 180 degrees

how fast can you make the satellite traverse a linear distance of exactly 1500 km (starting and stopping with a velocity of 0)? When answering , yous hould consider the following information:

-fully fueled, the satellite's mass is 2420 kg.

-the main thruster produces a constant force of F=96.8 N and consumes fuel at a rate of 1.6 grams/second.

-The rotational thrusters consume .8 g/s

-the satellite requires 400 seconds to complete a rotation of 180 degrees

- cannot use F=ma because in this scenario the satellites thrusters accelerate it by ejecting mass.Instead you should use a more general F= dp/dt, where p denotes momentum ( the product of mass and velocity, which depend on time)

-because the mass of the satellite at the beginning of the trip is not the same as the mass at toward the end, the duration of the first and last legs of this 1500 km trek will not be the same.

 

[gneill]

Satellites are objects that orbit another body, typically in circular or elliptical orbits. It's going to take some fancy work with the thrusters to perform a linear trajectory that starts and ends with zero velocity!

 

[Filip Larsen]

Please show us what equations you think are relevant to solving this problem, and try explain what you have already tried and where you are stuck. If you haven't been able to start solving this problem at all, I would suggest that you try break up the maneuver into segments such that you can attach some equations to each segment that when taken together will allow you to calculate the answer. You will most likely need to make assumptions along the way.

For instance, in the rotation part you know the satellite takes 400 seconds to rotate 180 degrees and it uses 0.8 g/s. Assuming the rotational truster are on for all 400 seconds (more or less half the time to increase rotational speed and the remaining time to decrease it back down to zero) you can write up equations that model how much fuel is used in that segment and, assuming constant speed along the 1500 m "track" during the rotational segment, how much distance it has covered in those 400 seconds.

 

[HallsofIvy]

If m is variable, then you are correct that "F= ma" cannot be used. However, the more general formula
F= \frac{dmv}{dt}= m\frac{dv}{dt}+ v\frac{dm}{dt}
can be used.