How can the spacecraft's orbit be determined with no thrust from its engines?

In summary: The 100,000 km is the radius. The 25,000 km is the distance from the surface of the earth. The line of momentum is the radius, so you want to use the 25,000 for the perpendicular distance.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)2T = m * 1.25x108Potential, V:V = -GMm r V = -m * 1.33x105Where M is the mass of the Earth
  • #1
Hoppa
38
0
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.
 
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  • #2
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
crap the powers and some of the symbols didnt turn out to well when i copied that over
 
  • #4
like i think i might have done somethign wrong, cause i don't 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
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 isn't 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
that's a thread killer !
 
  • #7
Pfft that's a "the poster disappeared"
 
  • #8
Kinetic, T:

T = 0.5m (1.5x104)2

T = m * 1.25x108
Your velocity is 4 km/sec, not 1.5x10^4.

V = -GMm
r
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.
Your radius is 100,000 km.

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

For circular orbit: (E = 0)
First statement is true. If you're using E=T+V, then second statement is false. Your orbit will be a parabola at E=0 (that's minimm escape velocity). E has to be <0 for all closed orbits, including circular.

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
If I understand the wording of the problem, this is where you want to use the 25,000 km (L=mrv*sin β ; β is the flight path angle relative to the radius).
 

1. What is a spacecraft orbit problem?

A spacecraft orbit problem refers to the mathematical calculations and analysis involved in determining and maintaining the trajectory of a spacecraft in orbit around a celestial body.

2. What factors affect the trajectory of a spacecraft in orbit?

The trajectory of a spacecraft in orbit is affected by multiple factors, including the gravitational pull of the celestial body it is orbiting, the spacecraft's velocity, and any external forces such as solar radiation and atmospheric drag.

3. How do scientists determine the optimal orbit for a spacecraft?

Scientists use mathematical models and simulations to determine the optimal orbit for a spacecraft based on its mission objectives, available resources, and various other factors such as the desired altitude and inclination of the orbit.

4. What challenges do scientists face in maintaining a spacecraft's orbit?

Maintaining a spacecraft's orbit can be challenging due to factors such as gravitational perturbations from other celestial bodies, atmospheric drag, and limited fuel reserves. These challenges require constant monitoring and adjustments to ensure the spacecraft stays on its desired trajectory.

5. How do scientists predict and prevent potential collisions between spacecraft in orbit?

Scientists use sophisticated tracking systems and predictive models to monitor the trajectories of spacecraft in orbit and identify potential collisions. To prevent collisions, they may adjust the orbits of the spacecraft to avoid any potential intersections or collisions.

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