What is astronaut 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. He has oxigen for 1 hour, and needs to get back to the ship.

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)?

Answer 1

The astronaut made some very simple calculations. The type you can make on a calculator. NASA, would have spent a couple of weeks trying to figure this one out.

The astronaut figured out where his space ship would be in one hour and threw the calculator in a direction exactly opposite of where the space ship would be. Since the space ship's orbit was circular with a period of 2 hours, it would be exactly on the other side of the earth. The astronaut therefore threw the calculator directly away from the earth.

Now both his intended orbit and the orbit of the spaceship are virtually circular. He can model his orbit as being 90 meters less. If he throws the calculator with enough momentum to give him a velocity in the direction of the earth such that he will cover 90 meters in one hour, that will be good enough. So he needs to throw the calculator at a velocity such that:

M1V1+M2V2 = 0

V2=-V1M1/M2

V1= 90m/hr = 0.025 m/s

V2= - .025 m/s *120/.2

V2= - 15m/s

Now, our boys from NASA are still trying to figure out the orbits. They realize that the astronaut's new orbit lies on the semi-latus r****m of an ellipse. They also need to figure out Kepler's laws and apply the laws to this situation. But they are having trouble inputting this data.

The NASA boys will figure it out the exact solution sometime tonight (As will I, so please leave the problem open). But by then, our astronaut will either be dead or alive and we can judge the intelligence of his choice. However, it is quite possible that the NASA boys will have figured out enough to have the astronaut make a minor correction to his orbit and jettison something else.

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Six minutes later, a young NASA engineer called up to the astronaut and told him to throw his 200g bag of M&Ms towards earth and remarkably the astronaut was saved.

You can get stuck worrying about Kepler and semi r****m latus rectums, but this young man realized the solution was much simpler. What is important is the perpendicular distance the spacecraft is ahead of the astronaut. This is:

V1= r sin (arclength/r) / ½ T

V1= 8080km sin (0.090km/8080 km) / 1 hour

V1= 1.571 m / hour

V1= 4.363e-4 m/s

ergo

V2=-V1*120/.2

= -.2618 m/s ...... slow drift

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Imagine a planet with a diameter of 8080 km. If you dropped a test particle down through a hole through the center of the planet, the particle would take 2 hrs to do an oscillation. Moreover, the velocity would follow a sinusoidal pattern.

Here it is as if the spaceship had dropped 1.57 m down the hole. For the astronaut to catch it, he only needs to make up this distance over the time period of ½ of an oscillation.

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Let's try a different approach. You still need to go from the semi latus r****m on one side to the one on the other side, both of which are R from the center of the earth and on opposite sides.  This means that you will cross the orbit of the spaceship one hour after the throw.  Hopefully, it will be there too.

Our astronaut is 90m behind the ship.  In order to catch the ship, he must be 180 m inside of the ship at 1/4 of the way around the orbit.  This is Kepler's second law of areas.  There is conservation of angular momentum.  Here is the derivation:

What is the area the spaceship has to traverse

A_ship = pi R^2/2 - .090/8080 * pi R^2

=pi R^2 (1/2 - .090/8080)

Note:  the term  .090/8080  the arc length that the ship is ahead of the astronaut.

A_astro= pi R * R min

A_astro =A_ship

Rmin= R *2 (1/2 - .090/8080)

Rmin =R - 180 m

[Also,

Rmin = R/1+e

e=R/Rmin -1

a= R/(1-e^2)

a= R*1.0000000004962969872542600785419

a is the semi-major axis.  The period for the astronaut will be 2hrs and 5e-6 secs]

Note:   throwing the calculator directly towards, or away from, the planet does not affect angular momentum.  This allows you to apply Kepler's second law to both orbits

Now, we can model this as a simple spring.  Period 2 hours, max displacement 180 m.

x = 180m * sin (wt)

w= (pi/1 hr)

v max = 180m pi/hr = .157 m/s

Now, for our calculator

V2= -.157m/s *600

V2= 94m/s

He needs to throw it away from the earth at 94m/s.  Roger Clemens on steroids couldn't even do that.