Why did I make this error?What is the error in my equation?

In summary, the problem is that I'm doing what seems to be perfectly reasonable algebra but I'm getting a result that makes no sense. I mistakenly assumed that momentum was equal to energy, which is not the case.
  • #1
JoAuSc
198
1
The problem is that I'm doing what seems to be perfectly reasonable algebra but I'm getting a result that makes no sense.

Let's say a photon travels from the left and hits a stationary solar sail which is perpendicular to the light. There is perfect reflection and a change in momenta for both objects. For momentum we have:

p_iphoton + p_iSS = p_fphoton + p_fSS =>
h/c*nu1 + m*(0)^2 = h/c*nu2 + mv
h/c*(nu1-nu2) = mv

energy:

KE_iphoton + KE_iSS = KE_fphoton + KE_fSS
h*nu1 + 0 = h*nu2 + mv^2
h*(nu1-nu2) = mv^2 = c*mv => v = c

(KE_iphoton is the photon's initial kinetic energy, p_iphoton is its initial momentum, KE_fphoton is the final KE, etc.; h = Planck's constant, c = speed of light, m = mass of solar sail, v = final velocity of solar sail, nu1 = initial frequency of photon, nu2 = final frequency of photon.)

Obviously, the final speed of the solar sail isn't the speed of light. What did I do wrong?
 
Physics news on Phys.org
  • #2
You did not take direction into account with your momentum conservation calculation.
 
  • #3
JoAuSc said:
The problem is that I'm doing what seems to be perfectly reasonable algebra but I'm getting a result that makes no sense.

Let's say a photon travels from the left and hits a stationary solar sail which is perpendicular to the light. There is perfect reflection and a change in momenta for both objects. For momentum we have:

p_iphoton + p_iSS = p_fphoton + p_fSS =>
h/c*nu1 + m*(0)^2 = h/c*nu2 + mv
h/c*(nu1-nu2) = mv
As Andrevh pointed out, the change in momentum is the sum of, not difference of, the magnitudes of the momentum before and after due to the 180 degree change in direction.

Also, you are using [itex]h/c\nu[/itex] for momentum of the photon which is not correct. Momentum of the photon is [itex]E/c = h\nu/c = h/\lambda[/itex].

energy:

KE_iphoton + KE_iSS = KE_fphoton + KE_fSS
h*nu1 + 0 = h*nu2 + mv^2
h*(nu1-nu2) = mv^2 = c*mv => v = c

(KE_iphoton is the photon's initial kinetic energy, p_iphoton is its initial momentum, KE_fphoton is the final KE, etc.; h = Planck's constant, c = speed of light, m = mass of solar sail, v = final velocity of solar sail, nu1 = initial frequency of photon, nu2 = final frequency of photon.)

Obviously, the final speed of the solar sail isn't the speed of light. What did I do wrong?
I don't follow your math here. You have to divide the change in momentum of the photon by the sail's mass to find the speed change in the sail. The m in your equation is the sail's mass not the mass of the photon.

[tex]\Delta P_{ph}/m_s = v_s = P_s/m_s[/tex]

[tex]KE_s = \frac{1}{2}m_sv_s^2 = P_s^2/2m_s[/tex]

AM
 
  • #4
Andrew Mason said:
As Andrevh pointed out, the change in momentum is the sum of, not difference of, the magnitudes of the momentum before and after due to the 180 degree change in direction.
Yes, I see it now. Thank you and Andrevh. I'm guessing you'd have a difference of frequencies in the energy equation, but a sum of frequencies in the momentum equation, so you can't substitute a change in photon energy with a change in photon momentum, at least not like I did.

Andrew Mason said:
Also, you are using [itex]h/c\nu[/itex] for momentum of the photon which is not correct. Momentum of the photon is [itex]E/c = h\nu/c = h/\lambda[/itex].
That's what I had, except that rather than clarify things with parentheses like I should have, I relied on order of operations. I used [itex]h/c*\nu = (h/c)*\nu = (h*\nu)/c.[/itex]

Andrew Mason said:
I don't follow your math here. You have to divide the change in momentum of the photon by the sail's mass to find the speed change in the sail. The m in your equation is the sail's mass not the mass of the photon.

[tex]\Delta P_{ph}/m_s = v_s = P_s/m_s[/tex]

[tex]KE_s = \frac{1}{2}m_sv_s^2 = P_s^2/2m_s[/tex]

AM
Let me clarify. In the last line of equations, I had
[tex]h*(\nu_1-\nu_2) = mv^2 = c*mv => v = c[/tex]
I believe the first two terms are, in fact, equal to each other, right? Now, based on my erroneous assumption that
[tex]\Delta P_{ph} = (h/c) * (\nu_1 - \nu_2)[/tex]
I got
[tex]\Delta P_{ph} = c * \Delta E_{ph}[/itex]
which implies that mv^2 = cmv.
 

What is a photon?

A photon is a fundamental particle of light and electromagnetic radiation. It has no mass, but carries energy and momentum.

What is a solar sail?

A solar sail is a spacecraft propulsion system that uses the pressure of sunlight to propel the spacecraft forward. It consists of a large, reflective sail that reflects photons from the sun to generate thrust.

How does a photon hit a solar sail?

When photons from the sun hit the reflective surface of the solar sail, they transfer momentum to the sail. This momentum is then used to propel the spacecraft forward.

How fast can a solar sail go?

The speed of a solar sail depends on the intensity of sunlight and the size of the sail. With current technology, solar sails can reach speeds of up to 100,000 miles per hour.

What are the potential applications of solar sails?

Solar sails are being considered for use in long-distance space travel and for exploring the outer reaches of our solar system. They could also potentially be used for satellite and debris deorbiting, as well as for powering future space probes and spacecraft.

Similar threads

Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
6
Views
5K
  • Introductory Physics Homework Help
Replies
1
Views
2K
  • Advanced Physics Homework Help
Replies
1
Views
2K
  • Special and General Relativity
Replies
1
Views
2K
  • Advanced Physics Homework Help
Replies
2
Views
6K
Back
Top