1/28/2014
Clues:
- The distance between LAX and BOS is 3,000 miles.
- Los Angeles is 3 hours behind.
- 100 km/hr = 60 mph
- Johannes Kepler and Newton determined that P^2 is proportional to a^3/M, where P is the period of a planet, a is the distance between the planet and the central mass, and M is the mass of the central body.
- Tool: a rock*
- Tool: a measuring cup half-filled with water*
* I had to run out for an interview before we actually got started on this, so I picked up a rock from outside. Then a buddy informed me I should use 2 g/cm3 as the density if I didn't have a scale or measuring cup. I did that, but if I hadn't, I would have calculated the volume of the rock by the amount of water displaced in the cup after adding the rock (knowing that 1 ml = 1 cubic centimeter of water). If I also had a scale to measure it's mass, the density of the Earth could be estimated as the ratio between the rock's mass and volume.
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| The rock I didn't end up using. |
Part I. Radius of the Earth
Convert the distance between Los Angeles and Boston to km.
We are given LAX-BOS: 3*103 miles. To convert this to kilometers, we use the relationship between the two implied by the speedometer:
100(km)/(hr) = 60(mi)/(hr) ⇒ 10km = 6mi ⇒ (5km)/(3mi) = 1
Using this relation to convert miles to km gives LAX-BOS: 3x103mi*(5km)/(3mi) = 5*103km
Estimate the circumference of the Earth based on the LAX-BOS distance and the LAX-BOS time difference
There are 24 time-zones if you circle the entire Earth. If I assume they are equally sized along latitudes, I can use proportionality to solve for the Earth’s circumference:
CE = ?km
(3hrs)/(5*103km) = (2.4*10hrs)/(CE)
CE3hrs = 2.4*10hrs*5*103km
CE = 8*5*103km = 4.0*109km
We can convert this to centimeters:
CE = 4*104km*(105cm)/(1km) = 4.0*109cm
Estimate the Earth’s radius from its circumference
The relationship between radius and circumference is
2πRE = CE
RE = (4.0*109cm)/(2π) ≈ 6.4*108cm
Part II. Mass of the Earth
Calculate estimate of volume of the Earth from knowledge of its radius.
The volume of a sphere relative to its radius, r, is (4)/(3)πr3. Since we have an estimate of the Earth’s radius, we can estimate it’s volume as
VE = (4)/(3)πR3Ecm3
Multiply volume*density to estimate ME, the mass of the Earth.
I’m using ρE = 2g ⁄ cm3 as the density of the Earth, based on my buddy’s estimate calculated in class. The mass of the Earth is
ME = VE*ρE = (4)/(3)πR3Ecm3*2(g)/(cm3)
ME = (8)/(3)πR3E(g)/(cm3)
ME ≈ 8R3E(g)/(cm3)
ME ≈ 8(6.4*108cm)3g ≈ 8*(6.4)3*1024 ≈ 8*2.5*1026
ME ≈ 1.7*1027g
Part III. Distance between Earth and Moon
Ask around for an equation representing Kepler’s Third Law:
P2 = (4π2a3)/(GM)
Where P is the orbital period, a is the distance from the central mass, G is the universal gravitational constant 6.7*10 − 8cm − 3g − 1s − 2, and M is the mass of the central body.
Kepler’s going to help us out here. Rearrange his equation to solve for a, the distance between the Earth and Moon.
P2GM = 4π2a3
(P2GM)/(4π2) = a3
((P2GM)/(4π2))(1)/(3) = a
Estimate P, the period of the Moon orbiting the Earth, as 3 x 10 days.
This is approximately the length of the lunar cycle - how long between us seeing the new moon to another new moon. In seconds:
3*10days(2.4*10hrs)/(1day)(6*10min)/(1hr)(6*10s)/(1min) = 3*2.4*6*6*104 = 7.2*3.6*105 ≈ 2.5*106s
Plug and chug, rounding as necessary to do math without a calculator.
a = (((2.5*106s)2(6.7*10 − 8cm3g − 1s − 2)(1.7*1027g))/(4π2))(1)/(3) = ((6*6.7*1.7*1012 − 8 + 27)/(4*9))(1)/(3)cm = ((6*6.7*1.7)/(3.6))(1)/(3)*1010cm
a = (6*2*1.7)(1)/(3)*1010cm = 22(1)/(3)*1010cm
a = 2.5*1010cm
Part IV. Radius of the Moon
This section will use knowledge that the width of a human thumb at arm’s length is approximately one degree. We approximate that the moon in the night sky takes up about one degree.
Method #1: Ratio relative to full orbiting distance of the Moon
Calculate the orbital distance of the Moon from the distance between the Earth and Moon centers, a, and the circumference of a circle equation.
Corbitmoon = 2πa
Now note that if the moon makes a full orbit around the Earth, it travels a distance relative to a stationary observer (on a stationary, non-rotating planet) on Earth of 360 degrees. We can set up the proportion between the distance of the Moon’s orbit and the Moon’s diameter, 2RM:
(Corbitmoon)/(360deg) = (2RM)/(0.5deg)
(0.5degCorbitmoon)/(360deg) = 2RM
(Corbitmoon)/(7.2*102) = 2RM
(2π(2.5*1010cm))/(7.2*102) = 2RM
(7.5)/(7.2)*108cm = RM
1.0*108cm = RM
Relative to the radius of the Earth:
RM = RE ⁄ 6.4 = 1.5*10 − 1RE
Method #2: Using triangle geometry
A second method we could use relies again on the approximation that the moon takes up 0.5 degrees in the sky from Earth. We also use the distance between the Earth and Moon, a.
tanθ = (RM)/(a)
a*tanθ = RM
We know that the side of the triangle should be a − RE, however, a dominates being two orders larger.
θ = 0.25deg
θ = 0.25deg(2π)/(360deg) = (0.25*6.28)/(3.60*102) = (1.5)/(3.6)*10 − 2 = 4.5*10 − 3
tanθ ≈ θ
for small enough angles.
tanθ ≈ 4.5*10 − 3
Subbing in the distance a
RM = (4.5*10 − 3)(2.5*1010cm) = 1.1*108cm
We get an answer very close to Method #1.
Part V. Mass of the Moon
To calculate the mass of the Moon, we can use the volume of the Moon.
VM = (4)/(3)πR3M
Where RM = 1.1*108cm, as calculated in the previous part. If we assume the Moon has the same density as the Earth; or, according to Prof. Johnson, the approximate density of cheese:
ρM = 2(g)/(cm3)
MM = VMρM
MM = ((4)/(3)πR3M)(2(g)/(cm3))
MM = 8R3Mg = 8(1.1*108cm)3(g)/(cm3)
MM = 8R3Mg = 8(1.3)*1024g
MM = 1.0*1023g
Relative to the mass of the Earth, ME = 1.7*1027g:
ME = 1.7*104MM
MM = (1.0)/(1.7)*10 − 4ME
MM = 0.6*10 − 4ME
MM = 6.0*10 − 3ME
