What was astronaut's solution?

Question

Small space module has circular orbit around the Earth
with period T = 2 hours. Astronaut in open space gear
(total mass M = 120 kg) is on exactly the same orbit
d = 90 m behind the module. After completing his
calculations, astronaut tosses his calculator (m = 200 g)
with certain velocity, and after time T/2 = 1 hour reaches
his module.

What was astronaut's solution (speed and direction of the toss)?

Edward,

While I share your belief that physics without equations becomes natural philosophy, I have a problem with your solution too.

As far as I can see you solve the problem in the frame of reference where the spacecraft is at rest. This frame of reference is non-inertial. Why do you assume that the velocity of astronaut will remain connstant in this frame of reference during the trip?

If you throw the calculator backwards,
then according to your own calculations the speed of astronaut will inrease by
3.6 E-6 = 2.5 cm/s, making this positon his new perigee. One hour later after 1/2 revolution the astronaut will find himself at new apogee, 90m * 2/pi = 60 meters higher, and further behind the module.

>> a=v^2/R

Nope. Centripetal acceleration relative to the module is
a = (Vo + dV)^2 / R - Vo^2 / R = 2Vo dV / R = 2 Omega dV (it is actually called coriolis force)

s = pi dV (T/2) = pi * d = 283m

This value is of course good only as rough estimate, because it is already not small compared to d
(the exact increase in altitude is 2d/pi).

Answer 1

Nice try Ed, but no.

This is kind of a bear of a problem.

My first thought was to toss the calc at the module, slow down, fall inward, and catch up on the other side of the orbit. Trouble is, the period goes like the semi-MAJOR axis, not the semi-MINOR one, so that doesn't help. A maneuver like the one I described requires a second burn (or calculator toss).

So I think you toss the calc outward and rather than fall inward, you push yourself inward. I don't feel like tackling the math (or reviewing orbital mechanics) to come up with the exact speed, but I think that's the gist of what he needs to do. This is kind of a risky maneuver, because if he does it wrong, his orbit not has a greater period and he falls farther and farther behind.

Edit--I just thought of a silly solution--not what your prof was looking for, but technically correct. Toss the flashlight directly forward with enough speed that the astronaut's velocity bounces back the other way. He'll go into a retrograde orbit with the same period and meet the shuttle an hour later on the other side. The only downside (besides not having enough arm strength) is that that flashlight is going to knock the hell out of his spacecraft, probably destroy it, and possibly knock it into a higher orbit (with a longer period), thus screwing up the rendez-vous. Still, if you can't figure the real solution, you can turn this in for partial credit style points.

Answer 2

Yes drift = 283m. Good point!
solution throw the calculator 12 times faster.
or/and compute the angle to compensate for radial drift .
Yes it is not that simple, nor is that complex. LOL
I should find the time to ponder it later. Thank you for the fun.
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Additional information for Alexander is way below...

To get back to the module that is 90 m away the astronaut tosses his calculator away from the module to catch the module.

P(before) = P(after) the momentum is taken relative to the module astronaut (m1, v1), calculator (m2, v2) system.
That is
0=m1v1 - m2v2
We have

v2=v1(m1/m2)
The speed of the astronaut is v1= s/t then we have

v2=[s/t](m1/m2)
v2 =[90/3600](120/.200)=15m/s apposite to the module's movement.
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To: Bekki B
Good try
You do the talk but can you do the walk?
Put down some equations to support you thoughts, and then do some calculations. We have to do computational physics not believes-based philosophy .

Second responce to Bekki B
Again you are not using calculations.
Lets use physics
Assuming a geocentric orbit with a period of 2 hours we have

We know that F (centripetal) = F (gravitational)
m w^2 R= G m M/R^2
M - Mass of Earth
G – Universal gravitational const.
R – Orbital radius
m - Mass of the satellite

R^3=GM/(w)^2

R = (GM/(w)^2)^(1/3)
GM= 398,600 km^3 per sec^2 for the Earth
w=2 pi/T=2 pi/(2 x 3600)=
w= 8.73 E-4 rad/sec

R=(398,600/(8.73 E-4)^2)^(1/3)= 8,060 km

V=wR= 8.73 E-4 8060 = 7.04 km/sec
Even at 15 ms/ sec fro the calculator the magnitude is too small to cause any problem for one orbital length.
In reality the speed of the astronaut is 15(.200/120)= 0.025 m/s or
%change = (.25 /7.04 E3)100%= 3.6 E-3 % its total tangential velocity.
Do you want to continue to split hair....?
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Alexander,
You being a bit hasty. The 2.5 cm/sec is a tangential velocity! Not centripetal.

The new centripetal acceleration (a) will be due to the additional tangential velocity of 2.5cm/sec

a=v^2/R
s=.5at^2
where t=3600 sec
s=.5[v^2/R](3600)^2
R= 8,060,000 m
Drift or s=0.5[(.025)^2 /8,060,000 ] (3600)^2=0. 0.0005m
So it is 0.05 cm not 60 m (unless I screwed up arithmetic somewhere)
Our astronaut is still safe!
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