Some of the Oldest Moon Rocks Found by the Apollo Missions Are Estimated to Be About 4.45 Billion Years Old

Astronomy 101 Problem Set #1 Solutions -- Fall 2005 10/2/19, 8)18 AM

Bucknell Unversity Astronomy 101 Problem Set #1 Solutions

Problem #1: Some of the oldest Moon rocks found by the Apollo missions are estimated to be about 4.45 billion years old. Calculate their age in seconds, and express the result in scientiﬁc notation (Use the relation 1 year = 365.25 days).

Solution: The tricky part about most problems usually lies in ﬁguring out how to apply what you already know to the problem at hand. In this case, we know lots of conversions between units of time (how many seconds in a minute, for example), but we don't know the requested conversion. So, let's start by writing down the conversions we know:

1 year = 365.25 days (remember leap year?) 1 day = 24 hours 1 hour = 60 minutes 1 minute = 60 seconds 1 class = 52 minutes

The last conversion is certainly correct, but it probably won't help us solve the problem at hand.

Now let's try to solve the problem in parts. We already know that 1 day = 24 hours, so how many hours are in 365.25 days ( = 1 year)?

1 day = 24 hours 365.25 days = 365.25 days x (24 hours / 1 day) = 8766 hours

For the next step, we do exactly the same thing, only this time, we make use of our knowledge of the conversion from hours to minutes. If there are 60 minutes in 1 hour, how many minutes are in 8766 hours?

1 hour = 60 minutes 8766 hours = 8766 hours x (60 minutes / 1 hour)

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= 525,960 minutes

The last step is yet another repeat of the same process, using our knowledge of the conversion from minutes to seconds. If there are 60 seconds in a minute, how many seconds are in 525,960 minutes?

1 minute = 60 seconds 525,960 minutes = 525,960 minutes x (60 seconds / 1 minute) = 31,557,600 seconds

Now, to ﬁnish it off, we multiply the number of seconds in a year by the number of years in the age of the oldest Moon rocks.

4,450,000,000 years x 31,557,600 seconds/year = 140,430,000,000,000,000 seconds

Now we have to express this answer in scientiﬁc notation. In order to do that, we need to express our answer as a number between 1 and 10 times a power of ten. To do that, we use the technique of moving the decimal point by noting that:

140,430,000,000,000,000 = 14,043,000,000,000,000 x 10 = 1,404,300,000,000,000 x 10 x 10 = 140,430,000,000,000 x 10 x 10 x 10 = 14,043,000,000,000 x 10 x 10 x 10 x 10 = 1,404,300,000,000 x 10 x 10 x 10 x 10 x 10 = 140,430,000,000 x 10 x 10 x 10 x 10 x 10 x 10 = 14,043,000,000 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 1,404,300,000 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 140,430,000 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 14,043,000 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 1,404,300 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 140,430 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 14,043 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 1,404.3 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 140.43 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 14.043 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 1.4043 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10

We can consolidate this number by noting that https://www.eg.bucknell.edu/physics/astronomy/astr101/prob_sets/ps1_soln.html Page 2 of 13 Astronomy 101 Problem Set #1 Solutions -- Fall 2005 10/2/19, 8)18 AM

10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 1017

and get for an answer

1.4043 x 1017 seconds

Note that we moved the decimal point by seventeen places in our number and multiplied by 1017. This is an easy way to convert big numbers to scientiﬁc notation without having to write down all those 10 x 10 x ... factors.

However, we're still not finished here. The last step of this part is to express our answer with the appropriate number of significant digits. Since you were given the age of the oldest Moon rocks as 4.45 x 109 years, and not 4.4 x 109 years or 4.4537809 x 109 years, we should assume that this number is only accurate to three significant digits. That is, the age of the oldest Moon rocks is somewhere between 4.445 x 109 years and 4.455 x 109 years old. Well, our translation of the age from years to seconds certainly hasn't added any accuracy to the result, so the answer should still be expressed with only three significant digits. The easiest way to do this is to just to truncate our decimal number from 1.4043 to 1.40.

So, our ﬁnal answer is 1.40 x 1017 seconds

Problem #2: For many of you, the metric system is fairly unfamiliar. Since ALL of science (and most commerce outside the US) makes use of the metric system, it would be a really good idea for you to have some comfort with the units. With that in mind, let's consider a few familiar objects and activities. For each question below, estimate the answer, and express that answer in metric units.

How tall are you? How far away is the LC from where you are right now (tell me where you are, too)? How fast do you walk? How fast do you drive on the highway?

Solution: For this problem, you'll likely come up with different answers, since you're not all the same height as me, and many of you probably drive pretty fast on the highways. Therefore, the grading on this problem reﬂects more about how you did the calculations, rather than the actual result. I'm 5 feet 11 inches tall, or a total of 71 inches. To express this height in metric units, we'll

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Bucknell Unversity Astronomy 101 Problem Set #6 Solutions

This Problem Set is due by 1:00 pm on Thursday, 13 October

Problem #1: Pluto's moon Charon has a radius of 635 km, and a mass of 1.8 x 1021 kg. What is its average density?

Based on this result, what kind of material do you think Charon is made of? (Hint: You can ﬁnd the densities of various materials on the 30 September web page.)

Solution: OK, we've got mass and radius, which is pretty much all we'll need to a density calculation. Recall that: density = mass/volume

The volume of Charon is: volume = 4/3 pi r3 = 4/3 x 3.14 x (635 km)3 = 1.07 x 109 km3

Note that the units of the radius were cubed along with the value, so our volume has units of cubic kilometers.

Now we can calculate the density:

density = mass/volume density = 1.8 x 1021 kg / 1.07 x 109 km3 density = 1.67 x 1012 kg/km3

Now this is a perfectly valid answer with perfectly valid units. Unfortunately, it's not in the same units as the table I gave you on the 30 September web page, nor is it in the units given in the back of the book. If you're interested in comparing this density to the ones I gave you, you'll need to convert:

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1.67 x 1012 kg/km3 x (1 km/1000 m)3 = 1.67 x 1012 x 1/109 kg/m3 = 1.67 x 103 kg/m3, or 1670 kg/m3

Note in this conversion I had to multiply by 1 km/1000 m three times to change the km3 units to m3.

Now this density lies between that of rock and water, so it might be reasonable to assume that Charon is made from a mixture of these. Really, what the average density tells you is that it's pretty unlikely that Charon has a large iron core, or that it's entirely rock and denser materials.

Charon is in fact an amalgam of rock and ice (it's really far away from the Sun, so the water is in the form of ice).

Problem #2: A 60-watt light bulb emits 60 joules/sec of energy ( 1 Joule/sec == 1 watt). Pretend for a moment that all of this energy is emitted in the form of photons with wavelength of 600 nm (this isn't true, of course -- as a blackbody, a light bulb emits photons of a wide range in wavelengths). Calculate how many photons per second are emitted by such a light bulb.

Solution: This problem asks you to calculate how many photons per second are emitted from a 60-watt lightbulb. Since 60 watts is 60 Joules per second, we know that we need enough photons to carry 60 Joules of energy each second.

So how much energy does one 600nm photon carry? We'll need to use the relation between energy and wavelength

Ephoton = h c / lambda = (6.63 x 10-34 J s) x (3.00 x 108 m/s)/ 600 x 10-9 m = 3.32 x 10-19 J

OK, so how many of these photons will we need to make 60 Joules? Let's assume we need some number N of them. Then,

60 Joules = N x 3.32 x 10-19 J

and

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A = 57.3o x (1.28 x 10-7) = 7.36 x 10-6 o

So the canals on Mars will appear to be 7.36 x 10-6 degrees wide.

Ground-based telescopes can only see features larger than 1 arcsec, or 1/3600th of a degree, which is 2.78 x 10-4 degrees. Since the canals we imagined above have an angular width more than 10 times smaller than this, they never could have been seen with ground-based telescopes.

In order to be seen by telescopes on the Earth, the canals would need to have an angular size of 1 arcsecond, or 2.78 x 10-4 degrees. We can ﬁgure out how big they'd need to be physically by using the Observer's Triangle again, but "backwards" this time.

2.78 x 10-4 o/57.3o = X / 7.78 x 107 km

where X is the physical size of the putative resolvable canal. Solving this equation for X, I get

X = 7.78 x 107 km x (2.78 x 10-4 o)/57.3o = 377 km

That's an absurdly wide canal -- about as wide as the distance from here to Ohio!

Problem #2: The Magellan spacecraft orbited Venus at an altitude of approximately 300 km (above its surface). Assume that the spacecraft's orbit was circular, and calculate its speed in its orbit. (Hint: You'll need to look up the mass and radius of Venus to answer this one.)

Solution: Here's a new twist on a speed calculation. In this case, you don't have a period, or any other time interval for that matter, so you can't use the reliable old speed = distance/time deﬁnition. This one requires a bit more thinking.

The Magellan spacecraft orbits Venus because of Venus' gravity. Without the gravitational pull from Venus, the spacecraft would ﬂy along in a straight line and wouldn't turn as it does in a circular orbit.

The acceleration that the spacecraft feels due to the pull of Venus is

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acceleration due to Venus' gravity = G Mass/radius2 = 6.67 x 10-11 m3/(s2 kg) x 4.87 x 1024 kg / (6.351 x 106 m)2 = 8.05 m/s2

where I looked up the mass of Venus in the back of our textbook, and I calcluated the radius of Magellan's orbit by adding together the radius of Venus and 300 km (converted to m).

Now, since I also know that the Magellan spacecraft travels in a circle, I know that speed, acceleration, and orbital radius are related as follows:

speed2 = acceleration x radius

So,

speed2 = 8.05 m/s2 x 6.351 x 106 m = 5.11 x 107 (m/s)2

and so, taking the square root, speed = 7150 m/s. That's about 16,000 mph.

Problem #3: Jupiter has an orbital period of 11.9 years.

a) Calculate the semi-major axis of its orbit.

b) Assuming that the orbits of both Jupiter and Earth are circular, calculate the minimum distance between these planets.

c) When Jupiter is as close to us as it ever gets, what is its angular size? (Hint: You'll need to look up Jupiter's radius for this one.)

Solution: OK, please tell me you know how to do part a) using Kepler's Third Law:

Period2 = semi-major axis3 11.9 year2 = semi-major axis3 141.6 = semi-major axis3

and, taking the cube root, I get that the semi-major axis = 5.2 A.U. Recall that this formulation

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