Measurements of the Speed of Light

[clairez93]

Homework Statement

The Apollo 11 astronauts set up a highly reflecting panel on the Moon's surface. The speed of light can be found by measuring the time it takes a laser beam to travel from Earth, reflect from the retroreflector, and return to Earth. If this interval is measured to be 2.51 s, what is the measured speed of light? Take the center-to-center distance from the Earth to the Moon to be 3.84 x 10^8 m and do not neglect the sizes of the Earth and the Moon.

Homework Equations

none given in the section

The Attempt at a Solution

I tried to simply to divide 3.84 x 10^8 m by 2.51, which got me the ansewr of 152988047.8 The book's answer is 299.5 Mm/s. I'm thinking my mistake is that I didn't do anything with the sizes of the Earth and the Moon, but I'm not sure how to incorporate those into this problem.

 

[Hootenanny]

Homework Statement

The Apollo 11 astronauts set up a highly reflecting panel on the Moon's surface. The speed of light can be found by measuring the time it takes a laser beam to travel from Earth, reflect from the retroreflector, and return to Earth. If this interval is measured to be 2.51 s, what is the measured speed of light? Take the center-to-center distance from the Earth to the Moon to be 3.84 x 10^8 m and do not neglect the sizes of the Earth and the Moon.

Homework Equations

none given in the section

The Attempt at a Solution

I tried to simply to divide 3.84 x 10^8 m by 2.51, which got me the ansewr of 152988047.8 The book's answer is 299.5 Mm/s. I'm thinking my mistake is that I didn't do anything with the sizes of the Earth and the Moon, but I'm not sure how to incorporate those into this problem.

That was indeed your mistake: The distance given is the distance from the centre of the Earth to the centre of the moon. Does the laser travel from the centre of the Earth to the centre of the moon?

 

[clairez93]

Hm, I guess not. So should I subtract the radii values of the sun and moon from the distance to get from surface to surface?
If I do that, I then get 375884500/2.51, which comes out to 149754780.9 m/s, which is still wrong, I believe.

 

[jeffreydk]

You've got it--I think you meant subtract the radii of the Earth and the moon though, not the sun. Think about it--velocity is distance over time--so you get...

$$c= 2\left(\frac{3.84 \times 10^8 - (E_R+m_R)}{2.51}\right)$$

 

[clairez93]

Oh, right. I forgot to multiply by 2.