# Equations of Motion with non-constant acceleration

## Homework Statement

Hi, I'm trying to work out the following equation that I have made for myself, based on the following information:
An object of mass m is travelling from a point r_0. The object has two forces acting on it, both of which are inversely proportional to the square of the distance from r_0.
I know this means that the ratio between the two forces is constant, and I have found the acceleration by using F=ma, with F being the difference between the two forces, divided by r^2.

How do I go about solving this equation for velocity at a certain distance r from r_0?

## Homework Equations

I know that (using simplified notation, lumping all constants together) a = (A-B)/mr^2 and that a = dv/dt, so we have a first order constant coefficient differential equation. However, it is the r^2 term that is causing issues, as this means it can't be solved easily.
Am I missing a trick here?

## The Attempt at a Solution

The reason why I wish to solve this is that this equation models the acceleration of a spaceship whose only forces acting upon it are the gravitational pull of the sun, and the radiation force acting on the sails of the spaceship. Both forces decrease like 1/r^2 as the ship moves away from the sun, hence the above equations. I am trying to find at what distance from the sun a particular velocity is reached, if at all. (As I imagine that if the difference between the two forces is not large enough, then this speed will never be attained)
As an aside, it is interesting to note that even if the radiation force is only slightly larger than gravity, then the spacecraft will eventually leave the solar system, and the sun's gravitational effect. But it will stop accelerating at some point, or the acceleration will become so small as to be negligable. But there have been no forces to slow it down, so it will be travelling at a great speed.

BruceW
Homework Helper
this isn't really homework, is it? Anyway, if you look at the form of your equation, what is the same and what has changed from the case when the object was acted on by only gravity?

Filip Larsen
Gold Member
You may also want to consider, that a solar sail for a "real" spacecraft really works best when it generates a force that points more or less in the direction of orbital travel around the sun, that is, a direction perpendicular to the direction towards the sun. By doing so, the "solar sail force" is not wasted against gravity but goes undiminished into raising the spacecrafts orbital energy and hence its distance from the sun.

If you only consider gravity and photon pressure, a quick back-of-the-envelope calculation gives that a perfectly reflecting solar sail that by some means have been brought to just "hang" a certain distance from the sun (that is, the sail has no transverse speed relative to the sun) will only accelerate straight out from the sun (that is, photon pressure will exceed gravitation) if the density of the sail is below around 1.5 g/m2 (1.5 gram per square meter). A density above that and gravity will be stronger than photon pressure and the sail would fall towards the sun.

It sort of is a coursework question, BruceW.
Anyway, I agree with you, Filip on the value of 1.5g/m^2 (unless the area of the sail is enormous, in which case it can, in theory, take any value).
This is where I am confused, as we are supposed to answer the question:

"...the spacecraft is realeased at a distance from the sun of 1AU with zero starting velocity. Determine the relations between it's total mass, sail area, and sail density for it to escape the sun's gravitational pull."

Would you take this to mean that it has no tangential velocity, therefore all acceleration would happen radially, as you say in the latter half of your post? If so, then I think I have answered the question....if not, then I'm going to have to have a drastic rethink.
Thanks.

ehild
Homework Helper
You need the speed at a given distance from sun, that is the function v(r)

Replace dv/dt with dv/dr dr/dt = dv/dr v. You get the separable equation
v dv/dr=K/r^2. (K=(A-B)/m).

This is the same as using conservation of energy.

ehild

Great. Thanks ehild. This then means that the velocity at any distance r from the sun of a spacecraft travelling radially outward, acted upon by gravity and radiation pressure is given by
v=√(2(B-A)/rm)

But this implies that for very large r, then the velocity will be small. But if it has been accelerating all that time, it will have accumulated a very large velocity....?

ehild
Homework Helper
Great. Thanks ehild. This then means that the velocity at any distance r from the sun of a spacecraft travelling radially outward, acted upon by gravity and radiation pressure is given by
v=√(2(B-A)/rm)

Do not forget the integration constant. The object starts from rest, but from a certain distance r0 from the sun, not from the centre! Also take care of the sign when you integrate 1/r^2.

ehild

Filip Larsen
Gold Member
...the spacecraft is realeased at a distance from the sun of 1AU with zero starting velocity. Determine the relations between it's total mass, sail area, and sail density for it to escape the sun's gravitational pull.

Would you take this to mean that it has no tangential velocity, therefore all acceleration would happen radially, as you say in the latter half of your post? If so, then I think I have answered the question....if not, then I'm going to have to have a drastic rethink.

Yes, from your quote of the question it does seems imply a zero tangential speed in which case the the relationship between the mentioned measures can easily be derived (without integration). You only need to integrate if you need speed or radial position as a function of time.