Determine the velocity of a package dropped from a satellite

[madsb2]

Homework Statement

A satellite with a total mass of M=450kg is moving parallel to the surface of the Earth. The satellite's velocity relative to the Earth is v_M=8km/s. A package with a mass of m=50kg is now ejected from the satellite, such that the package falls vertically towards the Earth's surface.

Determine the package's velocity relative to the satellite at the moment of ejection

Relevant Equations

My approach is based on the law of conservation of momentum
P=p_M*v_M+p_m*v_m
p=v*m

Momentum before ejection P_b:
P_b= (M+m)*v_M=8*500=4000, since the package is still on the satellite and isn't moving.

Momentum after ejection P_a:
according to the law of conservation of momentum P_b=P_a, and this gives me:
4000=450*8 + v_m*50 => (4000 - 3600)/50 = v_m = 8 km/s

But according to my answersheet the velocity for the package should be 9km/s

 

[DrClaude]

You seem to have calculated that nothing happens: both the satellite and the package are moving together at the same speed as before with respect to the ground.

I think the problem is badly worded, but here is how I make sense of it: the package is dropped such that it falls vertically means that it has 0 velocity parallel to the surface of Earth. Also, 450 kg appears to be the total mass before ejection, i.e., it includes the mass of the package.

 

[PeroK]

I think the problem is badly worded ...

Where do these problems come from? Anyone with any serious knowledge of physics would specify the reference frame in which the motion is "vertical", and say something about the Earth's rotation.

 

[kuruman]

4000=450*8 + v_m*50

The initial velocity of the package (assumed relative to the center of the Earth) is the same as the satellite, 8 km/s. That sets the forward direction for velocties. In what direction must the package be ejected (sign of v_m) so that its velocity is zero? Forward or backward?

 

[jbriggs444]

I think the problem is badly worded

Determine the package's velocity relative to the satellite at the moment of ejection

Indeed, if we imagine that the package and satellite velocities both change discontinuously at the moment of ejection then the relative velocity is technically undefined "at" the moment of ejection.

It would be well defined immediately after. Which I am sure was meant.

 

[nasu]

This may be just a fancy formulated question at the end of a chapter on collisions. In which case it is to be treated as a reversed completely inelleastic collision. Before the "collision" both object have a horizontal velocity (relative to the ground) of 8 km/h. After, the package has zero horizontal velocity. The satellite is just for fun. Can be a helicopter or a train. Not at that speed though.

 

[Tom.G]

I interpret "EJECTED" to mean that the package was pushed out of the satellite with some device, perhaps a spring.
This ejection imparted enough energy to the package to cancel its known orbital motion.

Now please keep two things in mind:
1) the momentum of the package was changed
2) for every action there is an equal and opposite reaction

What velocities are being asked?

Cheers,
Tom