•
•

## Help with math..

• Posted by a hidden member.
Ok, I know this might sound stupid to ask help for... but I'm kind of stumped.

Problem: Find an approximate value for the rate at which the moon orbits the earth. Assume that the moon's orbit is circular.

Hint: The average distance between the earth and the moon is 382,000,000 meters.

Ok, I know that the moon's orbit takes 27.3 days. So I found the diameter which would be 764,000,000. Then I found the circumference, which would be 764,000,000 x 3.14 = 2,398,960,000.

If D = RT, then is this how I would solve it?

2,398,960,000 = R x 27.3

2,398,960,000 divided by 27.3 = 87,873,992.67

So is the answer, 87,873,992.67 meters/day?
• Posted by a hidden member.
Your method is correct for an approximate. I'm getting 87,918,489 m/day, but close enough...
• Posted by a hidden member.
thanks man
• Posted by a hidden member.
Couple things:

1.) Use pi instead of 3.14

2.) If you need to use sig figs... your answer won't matter if you use pi or 3.14 since it is so large.

With 3 sig figs: 8.79E7 m/day

Without sig figs (rounded to nearest whole number): 87,918,564 m/day
• Posted by a hidden member.
first off, what is a sig fig? lol. I'm not good at math at all. And how can I use Pi instead of 3.14? I only have a basic kinda calculator, not one of those fancy smancy ones
• Posted by a hidden member.
hahah don't worry about significant digits then...If you have never heard of them then they are not important to your class.

pi ~= 3.1415926535898

May not seem like a big deal... but you were off by like 44,000+m/day

and round off onl at the end... not after each step
• Posted by a hidden member.
Oh ok, I got you now. Thanks man, appreciate it greatly!
• coolarmydude

Posts: 9196
The rate is velocity. You need to figure what meters/second (squared) is.
• Posted by a hidden member.
wouldn't that be a huuuuuge number?
• Posted by a hidden member.
Time is part of the SI system as well... since the data given was in meter & days , I would keep the answer in meter/days.

Basically, I'd say there is no reason to change the answer. Whats a better answer... 1 Liter or 1000mL? It is all relavent to what you are studying.

Chemical solutions = mL... Displacement of water from a battleship = L.

If I wanted the velocity of a racecar I wouldn't use m/day. I'd use m/s

With astronomy I'd use "bigger" units.

BTW m/s^2 is acceleration; m/s is velocity.
• Posted by a hidden member.
I'll just leave it with meters/day, like you said, since it's in the given information.
• coolarmydude

Posts: 9196
cjcscuba1984 said, "BTW m/s^2 is acceleration; m/s is velocity."

Correct! That's what I meant. Days will work too as it is a measurement of time. I just threw that out there as it is the physics equation for velocity. I wanted to mention velocity as the rate because I hadn't seen it mentioned.
• Posted by a hidden member.
coolarmydude saidcjcscuba1984 said, "BTW m/s^2 is acceleration; m/s is velocity."

Correct! That's what I meant. Days will work too as it is a measurement of time. I just threw that out there as it is the physics equation for velocity. I wanted to mention velocity as the rate because I hadn't seen it mentioned.

It's allllll coooooo maaaaaan. I'maa brian died meself right know....
• Posted by a hidden member.
While the information given is in days (which we don't actually know since he said "I know that the moon's orbit takes 27.3 days"), hardly ever is an answer in astronomy or physics given in terms of days. Units of seconds or years are more useful.

However, I'm sure that if it's for a class that as long as the answer is correct for a given unit system (and you make sure to say that you're doing it as meters/day), then the person grading it should have no reason to take off points.

Posts: 93
OK, so why is this TURNING ME ON? Brains must be sexier than I thought ....
• Posted by a hidden member.
tafkalil saidfirst off, what is a sig fig? lol. I'm not good at math at all. And how can I use Pi instead of 3.14? I only have a basic kinda calculator, not one of those fancy smancy ones

Three options for those without a good calculator:

1) If you're using Microsoft Windows, click the START button, click on Run, and type "calc". Go to the View menu and change it to Scientific. You will have buttons for pi, exponents, and other advanced functions. You can use Ctrl-C and Ctrl-V to copy/paste your numbers out/in to the calculator window.

2) Go to Google and search for "online scientific calculator". There are literally hundreds of free web-based calculators.

AND, Google will also do your unit conversions for you. Velocity is generally given in meters/second. Here's how you get Google to do the unit conversion.
• Posted by a hidden member.
the 'rate' could mean a couple of things...

1. the angular velocity... that is, the number of degrees of the arc of a complete circle (360º) covered in each second of motion by the moon

or..

2. the linear velocity... the linear rate at which an idealized point at the center of the moon moves in each second

1. angular velocity is the rate of change of angular displacement with time... let's use seconds. the moon moves a complete 360º circle in 27.3 days (according to your numbers).. so it sweeps through 360º in 2,358,720 seconds (27.3 days x 24hrs/day x 60mins/hr x 60secs/min)... for an average rate of 0.000152625 degrees/second. that's one measure of rate.

2. the linear velocity of a point on the surface of the moon is the angular velocity calculated in 1. times the radius of the circle, i.e. the average distance from the earth to the moon. in other words, velocity = (radius) x (angular velocity). so a point on the moon moves at an average linear rate of 0.000152625 x 382,000,000 meters or 58,302.81 meters/second.
• Posted by a hidden member.
Good point, veesam, that indeed the question could be interpreted to be either type of rate. However, the degree/second (or rather, it would be more useful to use arcseconds/second since the answer is so small, where there are 3600 arcseconds in a degree) rate would be useful if you wanted to observe the moon moving across the sky, whereas describing the general motion of the moon around the earth is better described by the meters/second (or other unit of time) result.
• Posted by a hidden member.
Wow, thanks guys. So much information. Math is my least favorite subject... I'm coming to y'all with all my problems, lol.
• Posted by a hidden member.
How would I convert 87,918,564 m/day into m/second? times it by 86,400?
• Posted by a hidden member.
no, divide by 86,400
• Posted by a hidden member.
1017.5759722222222222222222222222 meters/second?

Do I need to round this?

Sorry I sound so stupid lol, but I really do hate math.
• Posted by a hidden member.
That's why he was talking about significant digits earlier.
• Posted by a hidden member.