We're going to calculate how long it takes for rain to refill the Earth's oceans. We might want to know this number if, for instance, we were comparing Earth's oceans and Mars', or if we were looking at how long it would take water pollution on Earth to cycle its way through the entire water system (ocean -> cloud -> rain -> river -> ocean).
We can solve this problem (beyond guessing) by breaking it up into little pieces, each of which is simple enough by itself. We can then put the little pieces together and solve it the same way we calculate the number of sand grains on the Earth's beaches in class.
a. First, just for comparison, take a random guess at how many drops of rain fall in a year: ___________ drops. And, how long would it take for raindrops to fill up the Earth's entire oceans: __________ years. Don't worry about being accurate: just guess, and you can compare it to the number you'll calculate later on.
b. Let's start by calculating the amount of rain that falls on the
earth every year. A good place to start might be assuming it's about a
layer 1 meter deep - this might be wrong, but the amount of rain is
probably more than a centimeter, and less than 10 meters, so this is a
good place to start. (Feel free to pick your own value, too.) We can
multiply the area of the Earth (call it a sphere of radius 7000 km) by
the depth of the rain, and get the total volume of the rain that falls
every year:
Surface area of the Earth (written A_earth) = __6.2 1018 cm2.
Volume of rain every year (written V_year) = __6.2 1020 cm3/yr.
Convert units into seconds: V_second = __2 1013 cm3/s.
(You'll probably want to convert the units into cm before doing the math.)
(What you get for these particular values might vary if you use a layer different than 1 meter deep.)
c. How big is an average raindrop: 1 mm across? 3 mm? 1 cm? Pick
a value, and assume the raindrop is a sphere and calculate its volume:
V_drop = _________ cm3.
Now, divide the total volume
from above (V_year) by the volume of each raindrop, to calculate the
total number of raindrops that fall each year. Be sure your units are
right before dividing, since you can't divide mm3 by
km3 and get a meaningful number! It's probably easiest to
do everything in cm.
V_year / V_drop = _______________ rain drops in the ocean.
Using a 2 mm drop, I get V_drop = 10-2 cm3, and thus 6.2 1022 rain drops in the ocean. Your numbers may be different by an order of magnitude (factor of 10) or so, depending on what drop size you picked, and if you made any more assumptions.
d. What is the volume of the Earth's oceans? We can calculate this
based on assuming they cover perhaps a layer over the entire surface
about 10 km deep. Again, this number isn't 100% accurate, but the real
value is somewhere between 1 km and 100 km, so 10 km is a good guess -
or look up a better number. Of course, oceans don't cover completely
all of the Earth (only about 70%), but we'll assume for now that they
do. Using the same technique as in b), calculate the volume of the
oceans:
V_ocean = __6.2 1024 cm3, assuming 10 km deep.
e. Finally, let's divide the volume of the ocean by the rate at
which rain falls, and see how long it takes for rain to fill the
oceans:
V_ocean / V_year = __104 yr.
Were you surprised that it came out to a nice, round number? Why does it?
f. Does this number surprise you? Does it seem particularly long or short? How does it compare to what you guessed? Are there major assumptions that we made that are incorrect? Do you think the `real' value is larger or smaller than what we calculated here?
It seems long to me on the timescale of rivers and clouds (which can flood and empty in a matter of hours or days), but what it says is that the ocean is so much larger than either of these sources of water. Nevertheless, 104 years is a short time when compared with the age of the Earth (5 109 yr). It suggests that to have the Earth's oceans completely change -- or for Mars to lose or gain its oceans -- might actually be relatively quick. We've also ignored evaporation, so the actual dynamics would be quite different.
We made the assumptions that there's a meter of rain everywhere (perhaps too high?), that the oceans are uniformly deep, that raindrops are just one size, and so forth. We also ignored any water in the rivers. The net effect? It's probably a reasonable calculation.
Time = distance / speed = 1.5 AU / c = 2.25 1013 cm /
(3 1010 cm/s) = 750 s = 12 minutes
If it takes this long
for radio signals to get to Mars, by the time the rover driver would
have seen a rock in the way, it would be too late! 24 minutes too
late, since the radio signals would have to travel there and back.
b) What would be the delay time of a rover on the Moon, and could it be controlled directly by a person on Earth? On the moon, the delay time is 1.7 seconds each way, or 3.3 seconds roundtrip. Seems like a more reasonable rover to control in `real time'.
Let's say we sent rover to a planet in the Orion nebula, 450 ly away:
c) If our spacecraft traveled to Orion at one tenth the speed of light (written
as 0.1 c), how long would it take for it to reach Orion?
Hint: this is easier than it looks! How long would it take to go 450
ly if our spacecraft went at the speed of light? Half the
speed of light? Ten times the speed of light?
4500 yr; 450 yr; 900 yr; 45 yr. These results are simple, and require no calculations, if you remember the definition of a light year!
d) How long would it take for radio signals from Earth to reach Orion and return to us? Would this be a practical way to explore nearby planetary systems?
Since it would take 900 years for a roundtrip radio signal, this is dozens of generations of humans. Way too slow! If we want to interact with nearby planets, we might do better to find some closer, or visit them - neither of which are very simple, either!
If you give a clear, well-written argument using some facts to back up your case, I gave you some bonus points. This class is about astronomy, but if you can make a good case for astrology, I'll listen.
Last modified 13-Jun-2000