What is the final velocity of a sail after being hit by photons from a star?

In summary: You can eliminate it by setting ##v_i=0## and you should get back your first result.In summary, the luminosity of a star is defined by its energy output divided by the surface area of a sphere with radius r. When photons from the star hit a sail with mass M and cross sectional area A, the final velocity of the sail after being hit by photons for a short time t can be calculated using the equations for conservation of energy and momentum, assuming elastic collisions and that the photons reflect after hitting the sail. However, this calculation may not be accurate if the sail mass is small or if the sail is moving at relativistic speeds. In those cases, relativistic mechanics must be used to account for the energy and momentum
  • #1
Faiq
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16

Homework Statement


The luminosity of the star is defined as ##\phi = L/4\pi r^2##
The photons from the star hits the sail of mass ##M## and cross sectional area ##A##. What's the final velocity of the sail after getting hit by photons for a small time t? Assume the collision is elastic and the photons reflects after hitting the sail.

The Attempt at a Solution


##\delta p_s = 2\delta p_p= 2\frac{E}{c}=2\frac{LAt}{4\pi r^2c}##
##M\delta v = 2\frac{LAt}{4\pi r^2c}##
##v_f = 2\frac{LAt}{4M\pi r^2c} +v_i##

Is this correct? Because I assume the photons bounce back with a momentum of magnitude equal to initial but that's not stated anywhere.
 
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  • #2
It is correct as long as the sail mass is large and the sail is essentially at rest. Basically you are making the approximation that the sail rest frame is the CoM frame, in which the light will have the same frequency before and after the collision.

If the sail mass is small (same order as the energy hitting it during the time) you cannot neglect its recoil energy. If it is moving very fast you cannot neglect the Doppler shift and relativistic mechanics.
 
  • #3
I am in my early university classes and have only covered conservation of momentum and energy. Do you think I have to account for the things you mentioned? Btw the sail is not initally at rest. What should I do to account for that?
 
  • #4
Faiq said:
I am in my early university classes and have only covered conservation of momentum and energy. Do you think I have to account for the things you mentioned? Btw the sail is not initally at rest. What should I do to account for that?
But the thing is that you are not accounting for conservation of energy. (In your approximation he light has the same energy before and after and the sail has more after. It can only be neglected in the given limit.) Your approximation is only valid in the limits I told you.
 
  • #5
Yes that's why I was also concerned. Can you please suggest a way to account for the energy?
 
  • #6
Yes, make sure it is the same before and after. You have two unknowns and two equations, so it is only an algebraic exercise. If the sail is moving at non-relativistic speeds you can use the classical expressions for its momentum and energy. If not you need to use relativistic mechanics for both light and the sail.
 
  • #7
I am getting an equation like ##4E+M(v_i^2-v_f^2)=2Mc(v_f-v_i)## Not sure what to do now since I have to figure out ##v_i## in terms of remaining terms. (E is total photon energy emitted from star)
 
  • #8
Where did you get that from? Why don't you start from the beginning and write down the energy and momentum conservation equations (without assuming the light to have the same energy before and after).
 
  • #9
##\frac{E}{c}+Mv_i=-p_p+Mv_f##
##E+0.5mv_i^2=0.5mv_f^2+p_pc##
Solving both of them got me the equation that I mentioned above
 
  • #10
Looks fine to me. The initial speed ##v_i## is a collision parameter. Your result will of course depend on it.
 

FAQ: What is the final velocity of a sail after being hit by photons from a star?

What is momentum conservation?

Momentum conservation is a fundamental law of physics that states that the total momentum of a closed system remains constant over time, unless acted upon by an external force.

Why is momentum conservation important?

Momentum conservation is important because it allows us to predict the behavior of objects in motion and understand the effects of forces on those objects. It also helps us to analyze collisions and other interactions between objects.

How is momentum conserved in a closed system?

In a closed system, the total momentum before an interaction or event is equal to the total momentum after the interaction or event. This means that the sum of the momenta of all the objects in the system remains constant.

What are some real-world examples of momentum conservation?

Some examples of momentum conservation in everyday life include a child on a swing and a billiard ball colliding with another billiard ball. In both cases, the total momentum of the system remains constant before and after the interaction.

Can momentum be lost or gained in a closed system?

No, momentum cannot be lost or gained in a closed system. It can only be transferred from one object to another. This is known as the law of conservation of momentum.

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