Comet's acceleration due to ejected particles

[barbs00]

Homework Statement

Assume that water streams off a comet from the heated hemisphere at a velocity
vej . Show that, in the approximation that all ejected particles are directed towards the Sun, this causes an acceleration of the comet,
a=−3vej/r * dr/dt
where r is the radius of the comet.

Relevant Equations

vorb =√(GM/r)
a = dv/dt = dv/dr * dr/dt

What I did was consider a circular orbit around the Sun. We'll have the comet's velocity tangential to its movement (vorb) and a radial velocity pointing towards the sun due to the ejected particles (vej). Therefore we can find the total velocity to be vtot=√(vorb^2 + vej^2).
What I did after was dv/dr since we can say that a=dv/dr * dr/dt, however, this doesn't give me the right answer
Not sure what to do instead so I'd love some help!

Thanks x

 

[kuruman]

The "r" in a=−3vej/r * dr/dt is not the radius of the orbit; it the radius of the comet as stated in the problem.
Also, it is not clear whether $v_{ej}$ is the velocity relative to the Sun or relative to the comet. I would assume the latter, but I have not done the problem yet.

 

[collinsmark]

Hello @barbs00,

Welcome to PF!

Problem Statement: Assume that water streams off a comet from the heated hemisphere at a velocity
vej . Show that, in the approximation that all ejected particles are directed towards the Sun, this causes an acceleration of the comet,
a=−3vej/r * dr/dt
where r is the radius of the comet.
Relevant Equations: vorb =√(GM/r)
a = dv/dt = dv/dr * dr/dt

What I did was consider a circular orbit around the Sun. We'll have the comet's velocity tangential to its movement (vorb) and a radial velocity pointing towards the sun due to the ejected particles (vej). Therefore we can find the total velocity to be vtot=√(vorb^2 + vej^2).
What I did after was dv/dr since we can say that a=dv/dr * dr/dt, however, this doesn't give me the right answer
Not sure what to do instead so I'd love some help!

Thanks x

Don't worry about orbital velocity and orbital acceleration. Just ignore gravity altogether. Although considering orbital mechanics might ad a certain sense of realism, it is superfluous for this problem, and just adds unnecessary complexity.

Things you need for this problem:

Consider the comet as a perfect, solid sphere ("ball", if you're a mathematician) with uniform density, $\rho$.

The volume of a sphere is
$V = \frac{4}{3} \pi r^3$,

and thus it's mass is
$m = \frac{4}{3} \pi \rho r^3$

Force is the rate of change of momentum, and Newton's second law states that mass times acceleration is equal to force, thus,
$ma = F = \frac{d}{dt} \{ mv_c \}$
where $v_c$ is the velocity of the comet.

Now some questions for you:

Using the above equations, can you find an expression for the rate of change of the comet's mass?

Using conservation of momentum, can you find an expression for Newton's second law using the velocity of the exhaust instead of the velocity of the comet?

 

[haruspex]

Using conservation of momentum

Need to be careful here.
Momentum can only be taken to be conserved in a closed system, and in a closed system mass does not change. You can get away with this approach if the lost mass carries away no momentum in your frame of reference, but here it will.
Many regard writing $\frac{dp}{dt}=\dot mv+m\dot v$ where $\dot m$ is nonzero as simply invalid.

 

[collinsmark]

Need to be careful here.
Momentum can only be taken to be conserved in a closed system, and in a closed system mass does not change. You can get away with this approach if the lost mass carries away no momentum in your frame of reference, but here it will.
Many regard writing $\frac{dp}{dt}=\dot mv+m\dot v$ where $\dot m$ is nonzero as simply invalid.

Right, it's important to be careful.

If you include the momentum's of both the comet and exhaust then the system is closed. The lost mass (i.e., the mass of the exhaust) does carry momentum and that is a crucial part of the solution.

[Edit: Of course, not only are we ignoring gravitational forces/accelerations here but we also ignore the sunlight's radiation pressure. The only forces we're considering are the forces between the comet and the comet's own exhaust. So in this respect the system is closed.]

 

[collinsmark]

Need to be careful here.
Momentum can only be taken to be conserved in a closed system, and in a closed system mass does not change. You can get away with this approach if the lost mass carries away no momentum in your frame of reference, but here it will.
Many regard writing $\frac{dp}{dt}=\dot mv+m\dot v$ where $\dot m$ is nonzero as simply invalid.

And to be technically correct, I probably should have suggested invoking Newton's third law of motion, rather than conservation of momentum. But they're sort of the same thing here.

I guess my point is that if one can determine the rate of change of momentum of the comet's exhaust, one can indirectly determine the rate of change of momentum of the comet itself (or what remains of it).