Assume that the moon orbits the earth in a circular orbit. From the observed orbital period of 2

Assume that the moon orbits the earth in a circular orbit....

Key Concepts

Newton's Law of Universal Gravitation

This principle states that every pair of masses attracts each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. It is fundamental in celestial mechanics and allows us to calculate the gravitational force between the Earth and the Moon, which governs the Moon's orbital motion.

Centripetal Force in Circular Orbits

An object moving in a circular path requires a centripetal force directed towards the center of the circle to maintain its motion. In the context of celestial bodies like the Moon, the gravitational force provided by the Earth acts as the centripetal force, keeping the Moon in orbit. The balance between the gravitational force and the required centripetal force for circular motion forms the basis for determining orbital characteristics.

Orbital Period

The orbital period is the time it takes for an object to complete one full orbit around another object. This period is directly related to the distance from the central body and the gravitational force acting on the orbiting body. By measuring the orbital period and applying physical laws, such as equating gravitational force to centripetal force, one can calculate the radius of a circular orbit.

Transcript

00:01 In this question, we are asked to assume that the moon orbits the earth in a circular orbit and use the observed orbital period of 27 .3 days to calculate the distance between the moon and the center of the earth.

00:17 We are to assume that the moon's motion is determined solely by the gravitational force exerted on it by the earth and to use the mass given in appendix f.

00:26 So this is a fairly straightforward question once you list out your givens and you look at the formulas that you have available.

00:37 So let's go ahead with starting off with our given.

00:41 So in the question, we are given the moon's orbital period as 27 .3 days.

00:49 And we're also given the mass of the earth in the appendix f as 5 .97 times.

00:57 Times 10 to the 24 kilograms.

01:03 Other than that, that's pretty much all the information we have.

01:07 Now, we're asked to find the distance between the moon and the center of the earth.

01:13 That would be the orbital radius r.

01:16 And so it is tempting to take a look at this formula, which relates the orbital speed to the mass of the body, the major body, and the orbital radius.

01:29 The only problem is we don't have orbital speed, but we do have our orbital period.

01:37 So what we could do is take orbital speed and rate a formula for it in terms of the things that we do have and what we're trying to find.

01:48 So now we've got orbital speed is 2 pi r over the period.

01:53 That's basically just saying distance traveled, divided by time, that gives us speed.

02:01 And now we have a formula that includes only the radius, which is what we're trying to find, and things that we know.

02:11 What we need to do is kind of manipulate this a little bit.

02:15 We want to put all the r's together and everything else that's not an r on the outside.

02:21 So what i'm going to do is i'm going to rewrite the right -hand side here as gm to the one -half divided by r to the one -half.

02:33 That's exactly the same as what we had before.

02:36 It's just putting the square root on the top and the bottom separately and using an exponent instead of a square root symbol.

02:45 And i'll leave the left -hand side the same...

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