# two relativity problems

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• August 31st 2008, 08:55 PM
winterwyrm
two relativity problems
Thanks in advance, you guys are great!

Suppose that, while lying on a beach near the equator watching the Sun set over a calm ocean, you start a stopwatch just as the top of the Sun disappears. You then stand, elevating your eyes by a height of 1.70m, and stop the watch when the top of the Sun again disappears. If the elapsed time is t= 11.1s, what is the radius r of Earth?

and

A standard interior staircase has steps each with a rise (height) of 19cm, and a run (horizontal depth) of 23cm. For a particular staircase of total height 4.47m, how much farther into the room would the staircase extend if this change in run were made?
• September 1st 2008, 12:15 AM
CaptainBlack
Quote:

Originally Posted by winterwyrm
Thanks in advance, you guys are great!

Suppose that, while lying on a beach near the equator watching the Sun set over a calm ocean, you start a stopwatch just as the top of the Sun disappears. You then stand, elevating your eyes by a height of 1.70m, and stop the watch when the top of the Sun again disappears. If the elapsed time is t= 11.1s, what is the radius r of Earth?

Assume that the Sun is setting vertically, then its top edge is moving at 15 degrees per hour and in 11.1s it has moved:

$\theta = \frac{11.1 \times 15}{60 \times 60}$

degrees. This is the look down angle from your standing position to the apparent horizon.

Now you need a diagram, so see attachment.

Now we see that:

$
\cos(\theta)=\frac{r}{r+h}
$

RonL
• September 1st 2008, 12:18 AM
CaptainBlack
Quote:

Originally Posted by winterwyrm

A standard interior staircase has steps each with a rise (height) of 19cm, and a run (horizontal depth) of 23cm. For a particular staircase of total height 4.47m, how much farther into the room would the staircase extend if this change in run were made?

Part of this question is missing.

Number of steps is 4.37/0.19=23.

RonL
• September 1st 2008, 12:44 PM
winterwyrm
Thanks, but for the first problem, I keep getting that the answer is 1588.57m, this is likely not right, would it be possible if you could point me in the right direction for that one? Also, how did you calculate the degrees per hour?

For the other one, sorry, it is changing the stairs to a run of 28cm from 23cm, but I got this one based on what you gave me.
• September 1st 2008, 02:41 PM
CaptainBlack
[quote=winterwyrm;180721]Thanks, but for the first problem, I keep getting that the answer is 1588.57m, this is likely not right, would it be possible if you could point me in the right direction for that one? Also, how did you calculate the degrees per hour?[quote]

Well that is what I get, but the radius of the Earth is about 6000km so its wrong. I have checked the argument and I can't see anything wrong with it.

RonL
• September 1st 2008, 07:16 PM
winterwyrm
Thanks a ton, I'm sure it's just a trick question to throw off everyone who just googles the radius of the Earth or something, but how did you calculate the degrees per hour? Just wondering.
• September 1st 2008, 09:38 PM
CaptainBlack
Quote:

Originally Posted by winterwyrm
Thanks a ton, I'm sure it's just a trick question to throw off everyone who just googles the radius of the Earth or something, but how did you calculate the degrees per hour? Just wondering.

Well actually I didn't, being an astronomer I know it, but the Sun apparently moves 360 degrees in 24 hours which should give 15 degrees per hour.

RonL
• September 3rd 2008, 07:22 PM
winterwyrm
What if it's written like this?

Suppose that, while lying on a beach near the equator of a far-off planet watching the sun set over a calm ocean, you start a stopwatch just as the top of the sun disappears. You then stand, elevating your eyes by a height H = 1.16 m, and stop the watch when the top of the sun again disappears. If the elapsed time is t = 10.9 s, what is the radius r of the planet to two significant figures? Notice that duration of a solar day at the far-off planet is the same that is on Earth.
• September 3rd 2008, 09:05 PM
CaptainBlack
[quote=CaptainBlack;180771][quote=winterwyrm;180721]Thanks, but for the first problem, I keep getting that the answer is 1588.57m, this is likely not right, would it be possible if you could point me in the right direction for that one? Also, how did you calculate the degrees per hour?
Quote:

Well that is what I get, but the radius of the Earth is about 6000km so its wrong. I have checked the argument and I can't see anything wrong with it.

RonL
Well I now know what was wrong with this, its taking cos of an angle in degrees when in radian mode.

But now the angle is seen to be so small that we need series approximation to do the sums:

$\theta=0.000807\ {\rm rad}$

then:

$\cos(\theta)\approx 1-\frac{\theta^2}{2}$

and

$\frac{r}{r+h}=\frac{1}{1+h/r}\approx 1-h/r$

so we have:

$\theta^2 \approx \frac{2h}{r}$

so $r\approx 5220 \ {\rm km}$

(If you have sufficient prescission on your calculator you can do this without resort to approximation)

This is still too far from the true radius of the Earth for comfort. Refraction would tend to make the estimate larger (by up to 2000 km) rather than smaller. So I'm still not entirly happy with this.

RonL
• September 3rd 2008, 09:12 PM
CaptainBlack
Quote:

Originally Posted by winterwyrm
What if it's written like this?

Suppose that, while lying on a beach near the equator of a far-off planet watching the sun set over a calm ocean, you start a stopwatch just as the top of the sun disappears. You then stand, elevating your eyes by a height H = 1.16 m, and stop the watch when the top of the sun again disappears. If the elapsed time is t = 10.9 s, what is the radius r of the planet to two significant figures? Notice that duration of a solar day at the far-off planet is the same that is on Earth.

Same method will work

RonL
• September 3rd 2008, 10:00 PM
winterwyrm
how would you know the degrees per hour of this one though? I mean because it is a far-off planet this time, thanks for your tenacity, btw. I really appreciate everything you guys do on this site, I doubt I'd make my degree without this extra help.
• September 3rd 2008, 11:49 PM
CaptainBlack
Quote:

Originally Posted by winterwyrm
how would you know the degrees per hour of this one though? I mean because it is a far-off planet this time, thanks for your tenacity, btw. I really appreciate everything you guys do on this site, I doubt I'd make my degree without this extra help.

The apparent rate at which the star moves through the sky depends only on the day length, and you are told: "Notice that duration of a solar day at the far-off planet is the same that is on Earth". Hence 15 degrees per hour (or 15*pi/180 radians per hour).
• March 2nd 2009, 01:07 AM
uberjeep
Hi,

Could we assume that the earth is moving at a speed of 360deg/24hours? Or 1 degree per 240 seconds. Would this help with the calculation?
• March 2nd 2009, 04:33 AM
CaptainBlack
Quote:

Originally Posted by uberjeep
Hi,

Could we assume that the earth is moving at a speed of 360deg/24hours? Or 1 degree per 240 seconds. Would this help with the calculation?

Which is 15 degree/hour, which we already knew. (Which of course is only true on the celestial equator or if we are refering to longditude (or RA but that is usually measured in hours and miniutes, ..))

CB
• March 2nd 2009, 06:32 AM
uberjeep
This is what I've got:
H=1.70
Angular velocity (omega)=1 degree/240 seconds (from 360degrees/24 hrs). So the angle @ is 0.4625 degrees for T=11.1s

cos@ = r/r+h
therefore r = (cos@)(r+h)
r = rcos@ + hcos@
r-rcos@ = hcos@
r(1-cos@) = hcos@
r = (hcos@)/(1-cos@)
Substituting h=1.7m & @=0.4625, cos@ = 0.99999967420217
and r = 5217958.1618m
=5218km
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