Gravitation and Orbits help

In summary, the spacecraft has a jet engine that slows it down, and a solar sail that could help it leave the Solar System. If the power from the sun falls below that required to keep pushing the spacecraft further away, it may already have enough speed. However, the KE term might still be useful in the future.
  • #1
phyzz
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0

Homework Statement


A spacecraft orbiting the Sun uses its jet engine for slowing down its orbital rotation and changing the direction of its velocity. At the moment when the velocity is directed away from the Sun and has a magnitude of v = 30 km/s, the jet engine is switched off. At the same moment, the mass of the spacecraft is m = 1500 kg and the distance to the Sun is R = 150 x 10^9 m.

1) Find the distance at which the velocity of the spacecraft is zero
2) Now suppose that in addition to the jet engine, the spacecraft is equipped with a solar sail, which can be opened at the moment of time when the jet engine is switched off. A solar sail is a very large mirror, that reflects photons emitted by the Sun. The reflected photons change their momenta and thus generate a force directed away from the Sun: F = C/r^2 (where C = 1.2 x 10^17 Nm^2 is a constant and r is a distance to the Sun in meters). Is the power of the solar sail sufficient for this aircraft to leave the Solar system?
G = 6.67 x 10^-11, the mass of the Sun is M = 2.0 x 10^30kg


The Attempt at a Solution



1) I used F = ma
the force of gravitation being GmM/r^2 = ma

made a the subject of the formula

integrated to find v and used the information given to find the constant.

Then I integrated that v to get x, again finding a value for the constant using the relevant information.

I set the velocity equation equal to zero to find a value for t (ie the moment in time when the velocity is zero) and subbed that t into my x equation

I think my procedure is very coherent, at least it seems so to me, others are using the Conservation of Energy approach and now I'm getting really confused

2) I know that Power = work done/time

and that work done = (Force)(distance)

so I found the work done by the gravitation pull of the Sun by multiplying (GmM/x^2) by (x) where x is the distance I found previously and divided that by time (the same time I found when v = 0 previously)

I did exactly the same for the work done by the photons except this time I multiplied C/x^2 by x (x, again same as before) and divided by time (same as before).

I then verified the following inequality, if the power done by the gravitational pull of the sun > power done by photons then it doesn't leave the Solar System

I subbed in values for G and the M's and found that it indeed stays in the Solar System.

I really appreciate your comments and sorry if it's wordy!
 
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  • #2
phyzz said:
I think my procedure is very coherent, at least it seems so to me, others are using the Conservation of Energy approach and now I'm getting really confused
Should produce the same answer, but I would have thought using C of E would be much simpler.
if the power done by the gravitational pull of the sun > power done by photons then it doesn't leave the Solar System
Power is just the rate of energy transfer at some instant. Even if the power from the sun falls below that required to keep pushing the spacecraft further away, it may already have enough speed. Indeed, with a higher initial speed it might not have needed the sail at all.
OTOH, it's not immediately obvious that you can just look at the energy this time. The energy gained by the sail over time depends on how long it spent at different distances. It seems like you need to develop a differential equation for the motion. But as it happens, you don't. Look at the formula for the gravitational force at a given distance, and the one for the force on the sail. Notice anything?
 
  • #3
For part (1), conservation of energy is the way to go. I don't see how you could integrate the acceleration over time, since you have the acceleration in terms of distance (at least not easily -- show your work if you've done it!).

A very simply approach is to write the expression for the specific total mechanical energy of a body in orbit:

##\xi = \frac{v^2}{2} - \frac{\mu}{r}##

where ##\mu## is the gravitational parameter for the Sun, and is equal to ##G\,M_{sun}##.

In words it's the sum of the specific KE and PE, and is a constant of the motion of anybody in free-fall in a conservative field.

What happens to the KE term when the velocity goes to zero? Can you then find r?
 

What is gravitation?

Gravitation is the force of attraction between any two objects with mass. This force is responsible for keeping planets in orbit around their stars, and for the formation of galaxies and other large-scale structures in the universe.

How does gravitation affect orbits?

Gravitation is the force that keeps objects in orbit around a larger object, such as a planet or a star. This force is directly proportional to the mass of the objects and the distance between them. The larger the mass of the objects and the closer they are, the stronger the force of gravitation and the faster the orbit.

What factors influence the strength of gravitation?

The strength of gravitation is influenced by two main factors: the mass of the objects and the distance between them. The greater the mass of the objects, the stronger the force of gravitation. Similarly, the closer the objects are, the stronger the force of gravitation.

What is the difference between circular and elliptical orbits?

In a circular orbit, the distance between the two objects remains constant and the orbit is perfectly circular. In an elliptical orbit, the distance between the two objects varies and the orbit is more oval-shaped. This is because the force of gravitation is not constant throughout the orbit, causing the object to speed up and slow down at different points.

How can we calculate the orbit of an object using gravitation?

The orbit of an object can be calculated using Newton's law of universal gravitation, which states that the gravitational force between two objects is equal to the product of their masses divided by the square of the distance between them. By knowing the masses and distances of the objects, we can calculate the force of gravitation and use it to predict the orbit of the object.

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