Formula - History of Astronomy

Kepler's third law - shows the relationship between the period of an objects orbit and the average distance that it is from the thing it orbits. This can be used (in its general form) for anything naturally orbiting around any other thing.

Formula: **P ^{2}=ka^{3}** where:

- P = period of the orbit, measured in units of time
- a = average distance of the object, measured in units of distance
- k = constant, which has various values depending upon what the situation is, who P and a are measured.

There is a simplified version of this law: **P ^{2} = a^{3}** where:

- The object must be orbiting the Sun
- P = period of the orbit in years
- a = average distance of the object from the Sun in AU

**Typical Problems**

1. An object is orbiting around the star Gumby with a period of 80 years. If "k" = 2 (units of years and AU) in this system, what is the average distance of the object orbiting around Gumby in AUs?

Solution: You're trying to get "a" so you need to re-arrange the formula:

P^{2} = k a^{3}

a^{3} = P^{2} / k

a^{3} = (80)^{2}/2 = 3200

Now take the cube root of both sides

a = (3200)^{1/3} = **15 AU**

2. An object is orbiting around the planet Quackers with an average distance of 75,000 km. The value of "k" = 6.9 x 10^{-9} (units of days and km). What is the orbital period in days?

Solution: Plug into the formula

P^{2} = k a^{3}

P^{2} = 6.9 x 10^{-9} x (75,000)^{3}

P^{2} = 2.9 x 10^{6}

Take the square root of both sides

P = **1700 days**.

3. If an object has an orbital period of 85 years and is at an average distance of 24 AU from the object it orbits, what is the value of "k"?

Solution: Now you need to find the value of "k", so you need to rearrange the formula

P^{2} = k a^{3}

k = P^{2} / a^{3}

k = (85)^{2}/(24)^{3} = **0.52** (units are for AU and years).

4. If an object is orbiting the Sun with an orbital period of 15 years, what is its average distance from the Sun?

Solution: For this one you can use the "special" formula of Kepler's 3rd law -

P^{2} = a^{3}

a^{3} = (15)^{2} = 225

Take the cube root of both sides

a = (225)^{1/3} = **6.1 AU.**

5. If an object is orbiting the Sun at an average distance of 71 AU, what is its orbital period?

Solution: Use the "special" formula of Kepler's 3rd law -

P^{2} = a^{3}

P^{2} = (71)^{3} = 3.6 x 10^{5}

Take the square root of both sides

P = (3.6 x 10^{5})^{1/2} = **600 years**

Newton's Second Law of Motion = Force Law - This is the law that defines what force is. Since this is a physics class I am not going to have you use actual values in this law, but just relational values, so that you can determine how the parts of the formula change one one or more parts are changed.

Formula: **F = ma** where:

- F = force
- m = mass
- a = acceleration

**Typical Problems**

1. You have two cars stuck in the snow. One car (the blue one) has a mass that is 3 times greater than the other (the red one). You push them both with the same force, since you have only so much strength. How do their accelerations compare?

Solution: Look at the formula -

F = ma

In this case F doesn't change for the two situations, but m does and as a result so must "a". Think of it like a balancing act - but in this case the value of "ma" has to balance for both cars. So you have ma for the red car and ma for the blue car. And they have to be equal one another (since you are exerting the same force on both of them). Therefore

ma = ma

But m is 3 times greater than m

So a must be 3 times greater than a for both sides to balance.

Or you could think of it this way - with size representing differences
ma = ma

2. Two asteroids with the same mass are moving through space. Asteroid Beatles is accelerating 6 times more than asteroid Stones. Both will hit the Moon. Which will have a greater force of impact and how much greater will it be?

Solution: Look at the formula -

F = ma

Since "m" is the same for both object, we only need to look at "a". It is 6 times greater for the Beatles, so the force that exerted by it is 6 times greater than the force exerted by asteroid Stones. That's all there is to it.

Newton's Universal Law of Gravitation - The most important law in the Universe. This one you really have to know and know how it is changed by different circumstances. And like the previous law, we won't be plugging in actual values in general, only when it is vital to a problem, which isn't very often.

Formula: **F = G M _{1}M_{2}/R^{2}** where:

- F = force of gravity
- M
_{1},M_{2}= masses of the objects involved - R = distance between their centers of mass (usually just their centers)
- G = a constant.

**Typical Problems**

1. If you are on the surface of the Earth what does the force of gravity depend upon in that situation?

Solution: Look at the formula -

F = G M_{1}M_{2}/R^{2}

In this case there are two objects involved, you and the Earth. So one of the masses is your mass and the other is the mass of the Earth. What is R? That is the distance between the objects, in this case the distance from your center to the center of the Earth. For the most part the distance to your center doesn't matter much especially when compared to the size of the Earth. So the formula can be rewritten as

**F = G M _{you}M_{Earth}/R_{Earth}^{2}**

2. What happens to the force of gravity that you feel if you go on top of a very high mountain?

Solution: Look at the formula we got from the previous question -

F = G M_{you}M_{Earth}/R_{Earth}^{2}

The only thing that is going to change is the "R_{Earth}" value - it will get bigger as you get further from the center of the Earth. It is also on the bottom of the formula, that means you'll be dividing by a larger number, and the value gets squared, which makes it even larger. The over all effect is that as "R" goes up, the force of gravity must go down. **So you would feel less of a pull from the Earth if you go up a very high mountain**.

3. You travel to the planet Vulcan, which has a mass that is 4 times greater than the Earth's and a radius that is 3 times greater. How does the force of gravity on Vulcan compare to that of the Earth?

Solution: Look at the formula we got from the first question in this section-

F = G M_{you}M_{Earth}/R_{Earth}^{2}

By going to another planet you are going to change the contribution of the value of the mass and radius of the planet.
In this case you are going to increase the mass by a factor of 4, so the top of the formula is getting multiplied by 4. Also the radius is 3 times greater, so the bottom value of "R" gets multiplied by 3 - but wait! The radius value is squared, so if you multiply the radius by 3, the end result is a change of 3 x 3 = 9. So the bottom of the formula gets multiplied by a factor of 9.

What's the end result? The top is multiplied by 4, the bottom by 9, so the gravity is 4/9 = 0.44 of the Earth's. Or put another way it is **4/9 that of the Earth's**, or 44% of the Earth's gravity. Even though the mass of the planet is greater, the gravity goes down!

4. You travel to the planet Mirana, which has a mass that is 2 times smaller than the Earth's and a radius that is 4 times smaller. How does the force of gravity on Mirana compare to that of the Earth?

Solution: Look at the formula we got from the first question in this section-

F = G M_{you}M_{Earth}/R_{Earth}^{2}

By going to another planet you are going to change the contribution of the value of the mass and radius of the planet.
In this case you are going to decrease the mass by a factor of 2 (cut it in half), so the top of the formula is getting multiplied by 1/2. Also the radius is 4 times smaller, so the bottom value of "R" gets multiplied by 1/4, but that has to be squared. So the bottom changes by a factor of 1/4 x 1/4 = 1/16.

What's the end result? The top is multiplied by 1/2, the bottom by 1/16, so the gravity is (1/2)/(1/16) = 16/2 - remember, dividing by a fraction is the same as multiplying by its reciprocal. So the force of gravity on Mirana is 16/2 = **8 times greater than the Earth's**.