SIMG-217-20032
Solution Set #1
1. The planet Mars is simultaneously photographed from two points on the equator of
Earth — one where Mars is observed to be rising and the other where Mars is setting.
The photographs show that the position of Mars against the background stars differ
by 41" (arcseconds). You need to find a measurement of the distance to Mars.
This is another version of the parallax measurement that was mentioned in class and
considered in the notes on “angles, distances, and measurements”
(a) Draw a diagram of the positions of the telescopes, Earth, and Mars when the
observations were made.
(b) Sketch two possible images of Mars that satisfy the data given.
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(c) What additional information do you need to find the distance to Mars?
In the sketch for part (a), we have a triangle with sides x,y,r, where x is the
distance from the center of Earth to the center of Mars, r is the distance from the
surface of Earth to the center of Mars, and y is the distance from the center of
Earth to the surface (i.e., the radius of the Earth). We know two of the angles
(the “skinny” angle that is 41”
= 20.5”, and the right angle at the intersection of
2
sides x and y). From trigonometry, we can see that:
tan [20.5”] =
or:
x=
y
x
y
tan [20.5”]
Thus we can find x if we know y, the radius of the Earth which we can get from
the diameter at the equator.
(d) Find the information required in part (c) and compute the distance to Mars in
kilometers and astronomical units.
I looked up the diameter of the Earth on the internet at:
http://geography.about.com/library/faq/blqzdiameter.htm
Diameter of Earth = 7, 926.41 miles = 12, 756.32 kilometers
8000 miles is close enough for our purposes, which translates to 12,875 km. Again,
for our purposes, 12,900 km is close enough. Thus the radius of the Earth is about
6,450 km, so we can substitute into the equation for x:
x=
6450 km
tan [20.5”]
But we’re not done yet — how to deal with the tangent of the angle? You could
just enter this into some calculators, but most require "fractional angles". I’ll do
the preferred way first, based on the conversion between arcseconds and radians.
Recall that:
1 radian ' 206265 arcsec
or
1 arcsec '
1
1
radians '
radians = 5 × 10−6 radians = 5µradians
206265
200, 000
Thus the angle of 20.5" is approximately equal to:
¶
µ
−6 radians
= 102.5 × 10−6 radians
20.5 arcsec ' 20.5 arcsec × 5 × 10
arcsec
−4
= 1.025 × 10 radians ' 10−4 radians
Now recall the relation between the angle, its sine, and its tangent for small angles:
sin [θ] ' tan [θ] ' θ for small θ, i.e., θ <
2
π
radians
10
Thus we can say that:
¤
£
tan [20.5 arcsec] ' tan 10−4 radians ' 10−4
Substitute this into the equation for x:
x =
¡
¢
6450 km
6450 km
+4
3
=
6450
×
10
km
=
6.45
×
10
'
× 104 km
tan [20.5”]
10−4
= 6.45 × 107 km ' 65 × 106 kilometers
Now we need to convert this distance to astronomical units, where 1 AU is the
distance from the Earth to the Sun. We know that this is approximately 93,000,000
miles. I typed “Astronomical Unit” in Google and up popped the value:
1 Astronomical Unit = 149, 598, 000, 000 meters
= 149, 598, 000 kilometers
' 150 × 106 kilometers
So the distance to Mars in A.U. is approximately:
65 × 106 kilometers
65
=
A.U. ' 0.43 A.U.
kilometers
6
150
150 × 10
A.U.
which says that Mars was a bit less than half as far away from Earth as the Sun.
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2. At the same time as the observations in problem #1, a radio telescope is aimed at
Mars and a pulse of radio waves is transmitted in that direction. Exactly 5 minutes
and 8 seconds later, a signal is received by the telescope. The telescope operator says
that this is the echo of the radio pulse reflected back from Mars. Agree or disagree
with the statement and explain your reasons.
This requires us to find the distance that the radio wave traveled in 5 minutes and 8
seconds, or 308 seconds. The velocity of a radio wave is the same as a light wave. In
vacuum, it travels at c, which is 186,000 miles per second, or (better still)
c = 2.997 × 108
meter
meter
kilometers
' 3 × 108
= 3 × 105
sec
sec
sec
In 308 seconds, the radio wave travels:
kilometers
= 924 × 105 kilometers
sec
= 9.24 × 107 kilometers = 92.4 × 106 kilometers
' 308 sec ×3 × 105
BUT, this is the “round-trip” time for the light to go out and return. So the actual
distance to the reflecting object is half of this distance:
distance to reflecting object =
2
' 46.2 × 106 kilometers < 65 × 106 kilometers
In other words, the measured radio wave either reflected from an object that is closer
to Earth than Mars, or came for something else. We may have to put the X-Files crew
on this case....
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3. I stated in the lecture that the “rule of thumb” for the eye’s resolution is 1 minute of
arc. We calculated that the limiting resolution of a telescope whose lens (or mirror)
diameter is 3.5 inches is 1.2 arcseconds. Assume that these measurements are made at
the same wavelength λ and use them to determine the diameter of the lens of the eye.
We know that:
λ
,
D
where λ is the wavelength of light and D is the diameter of the optical system. In this
case, we have two different ∆θ0 s and two different D’s:
∆θ '
λ1
=⇒ λ1 = D1 × (∆θ)1
D1
λ2
'
=⇒ λ2 = D2 × (∆θ)2
D2
(∆θ)1 '
(∆θ)2
If we assume that λ1 = λ2 (which is very reasonable, since we likely would measure the
resolution at the same wavelength), then we have:
D1 × (∆θ)1 = D2 × (∆θ)2
If D 1 is the diameter of the telescope and (∆θ)1 is the corresponding resolution, and
D 2 and (∆θ)2 are the same for the eye, we can rearrange this equation to obtain:
(∆θ)1
1.2 arcsec
1.2 arcsec
= 3.5 inches ×
= 3.5 inches ×
(∆θ)2
1arc min
60 arcsec
1
= 3.5 inches ×
= 0.07 inches ' 1.8mm
50
D2 = D1 ×
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4. (BONUS PROBLEM — EXTRA CREDIT) The Moon’s position among the stars is
observed at moonrise and at the moment the moon reaches the point directly overhead,
which happens 6 hours and 12 minutes later. The position of the Moon on the field
of background stars shifted to the east by 148´ of arc. From a table in an almanac,
it is known that the Moon moved eastward on the celestial sphere by 33´ per hour.
Use these data to find the distance to the Moon and compare it to the approximate
distance of one quarter of a million miles.
See sketch first:
6 hours and 12 minutes is:
6+
12
hours = 6.2 hours
60
at 33 arcminutes per hour, the Moon would move
6.2 × 33 min = 204.6 min
BUT the observed motion is only 148 arcminutes, so the Moon moved about 56.6 arcminutes LESS than it should have. The difference between these is due to the Earth’s
rotation during this time. From the drawing below, we see a triangle with sides r and
x and the angle of 56.60 .
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The solution for the unknown distance x is:
tan [56.60 ] =
r
r
=⇒ x =
x
tan [56.60 ]
But r is the radius of the Earth, which we already found to be approximately 6,450 km. The
angle of 56.60 is just less than a degree. In radians, its size is:
θ = 56.60 =
56.60
2π radians
' 0.943◦ = 0.943 ×
' 0.016 radians
0
60 per degree
360◦
But tan [0.016] ' 0.016, so we have:
x '
=
6450 km
6450
'
km ' 4.031 × 105 km
tan [0.016]
0.016
403, 100 km ' 250, 528 miles
Just about right! From:
http://starryskies.com/The_sky/events/lunar-2003/eclipse8.html
the mean distance is about 238,712 miles and the greatest distance is 252,586 miles.
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