[07199534]

I am trying to find out an approximate value for the rate at which the moon orbits the So what I did is 2 * 3.14= 6.28 then the distance between the earth and moon 382,000,000. So 6.28 * 382,000,000 = 2,398,960,000 then I divided 2,398,960,000 by 27.3 which is the amount of days it takes the moon to orbit earth. My answer is 87,873,992.67 meters per day. I am hoping someone can tell me if this is the correct way to do it and the correct answer.

[galactus]

The moon's orbit is elliptical, not circular. But you are close. It should be a larger than that because of the elliptical orbit as opposed to circular.

One way you could look at it is that the mean orbital speed of the moon is

1023 meters/second. There are 86,400 seconds in a day.

Therefore, (86400)(1023)=88,387,200 meters/day.

[Jamaine]

find an approximate value for the rate at which the moon orbits the earth

[Jamaine]

3.14=6.8 382,000,000 *6.28= 2,398,960,000

[galactus]

find an approximate value for the rate at which the moon orbits the earth

For rate, what sort of units do you want? Miles/hour? From our answer above, we can convert meters to miles and then divide by 24. This gives us 2288.39 miles per hour the moon travels in its orbit.

[lilmaninsc]

I had the same question in my math class.

Ok, first off you need to get the diameter, which would be 764,000,000.

Circumference is 764,000,000 Pi, you want to use the approximate value of Pi or you could be off by a couple thousand. C = 240,017,687.34

Since D=RT, 240,017,687.34 = R times 27.3, divide by 27.3 to get R by itself.

240,017,687.34 divided by 27.3 = 87,918,563.64 meters/day

The units used is normally meters/second, so simplify more.

There are 86,400 seconds in a day.

So, 87,918,563.64 divided by 86,400 = 1,017.58

Answer should be, 1,017.58 meters/second