Spacecraft orbit problem

1. May 29, 2005

Hoppa

this is the problem which i need to solve:

A space-craft is at a distance of 100; 000 km from the Earth’s centre and is moving
with a speed of 4 km/s, as measured in the assumed inertial frame of the Earth, with no
thrust from its engines. It is currently moving along a direction that passes the Earth
by a perpendicular distance of 25,000 km from the Earth’s centre.

1.The space-craft continues to travel in the gravitational field of the Earth with
no thrust from its engines. Calculate the parameters associated with the spacecraft’s
orbit. What is the closest distance between the space-craft and the Earth’s
centre? Will it hit the Earth? If not, will the orbit cause it to return to the Earth
again?

2. Suppose the space-craft fires its engines in a radial direction with respect to the
Earth’s centre to try to get into a circular orbit about the Earth. Could this be successful
and, if so, at what radius would it orbit the Earth? Neglect any reduction
in mass of the space-craft on firing its engines.

2. May 29, 2005

Hoppa

this is what i have got:

Spacecraft distance of 100,000km
Speed of 4km/s
Perpendicular distance of 25,000km

Let the mass of the asteroid be m. Its angular momentum, with respect to the centre of force, is a constant and expressed as

L = r * mv

This can be evaluated using values for the spacecraft as originally observed. The magnitude is the product of the momentum by the perpendicular distance between the line of the momentum and the centre of the Earth.

L = m * 1.0x105

The energy of the spaceship is also a constant and can be evaluated from the components, kinetic, T, and potential, V, as originally observed.

Kinetic, T:

T = 0.5m (1.5x104)2

T = m * 1.25x108

Potential, V:

V = -GMm
r
V = -m * 1.33x105

Where M is the mass of the Earth.
These yield:

E = m * 1.124x108

The distance of closest approach can be found from the solutions of the effective potential at the particular energy. The effective potential is:

Veff = -G Mm + L2
R 2mr2

And the distance of closest approach is given by:
.
1 = m2MG + Ö æm2MGö2 + 2mE
r L2 è L2 ø L2

Substituting all the values, with m canceling throughout.

.
1 = 5.98x1024 * 6.67x10-11 + Ö æ5.98x1024 * 6.67x10-11 ö2 + 1.124x108
r 1.0x1010 è 1.0x1010 ø 1.0x1010
.
= 3.98860x1024 + Ö 1.59x1049 + 1.124x1018
.
= 3.98860x1024 + Ö 3.18x1049

= 3.98860x1024 + 5.64x1024

= 9.63x1024

Since (E > 0) then the orbit is unbounded.

For circular orbit: (E = 0)

E = V0 = -mk2
2L2

3. May 29, 2005

Hoppa

crap the powers and some of the symbols didnt turn out to well when i copied that over

4. May 29, 2005

Hoppa

like i think i might have done somethign wrong, cause i dont know how to finish it off, with like what the cloest distance to earth will be and whether it would hit the earth or not.

5. May 30, 2005

whozum

Thats not very easy to read but I did notice a few things,
First off, your pulling some things out of your ass here,
Kinetic, T:

T = 0.5m (1.5x104)2

Where did you get v = 1.5e4? This isnt given anywhere in the problem.

Second off, the speeds and distances are given in kilometers, its best to convert to meters, or you could just keep them and keep track of your units so as not to confuse your final result.

6. May 31, 2005

pete worthington

that's a thread killer !

7. Jun 1, 2005

whozum

Pfft thats a "the poster disappeared"

8. Jun 1, 2005

BobG

Your velocity is 4 km/sec, not 1.5x10^4.

Your potential is -GMm/r.
You mixed meters and kilometers together and/or did something horrible with your powers of ten. GM is 3.986 x 10^14 in meters; 3.986 x 10^5 in km. Since all your other measurements are in km, use the kilometers.