Speaking of physics.....quick homework question

[sandmangeck]

I need to know how to get the answer.
Here's the question.....
The period of the earth around the sun is 1 year and it's distance is 150 milllion km from the sun. An asteroid in a circular orbit around the sun is at a distance 273 million km from the sun.
What is the period of the asteroid's orbit?
The answer is 2.45531 years.

GO!

[Muaguana]

The time period of uniform circular motion is related to its radius and velocity: T=(2*pi*r)/v, where r is the radius of the orbit and v is the velocity of the asteroid. You can calculate the velocity by the equation v=(G*M/r)^1/2, where G is the gravitational constant and M is the mass of the sun.

Edit: It seems odd that they would provide the answer to six significant figures when the most significant figures used for the data is three. I thought proper sig figs was the first thing they taught in Physics?

[Time Monkey]

what?

[Muaguana]

Originally Posted by Time Monkey

what?

Step 1. Look at equations.
Step 2. Solve for unknown variables.

Very simple, really. System of two equations with two unknowns. Plug and chug the numbers, and there's your answer.

Edit: Here's the composite equation to get the answer in one swing: T=(2*pi*r)/((G*M/r)^1/2), where T is the period, r is the distance from the asteroid to the sun (radius), G is the gravitational constant, and M is the mass of the sun. You might want to use SI units for distance and time, since the gravitational constant is in meters cubed per kilogram per seconds squared. Just take the answer in seconds you get and convert it to years (use the conversion factor 1 second = 3.16888 × 10^-8 year).

[T-Bird]

Hi.

Originally Posted by Muaguana

Edit: It seems odd that they would provide the answer to six significant figures when the most significant figures used for the data is three. I thought proper sig figs was the first thing they taught in Physics?

My educated guess would be that they require the students to use at least one constant (or to came up with the result) that has six significant figures to score full points for the exercise.

As we know, in science the "correct" answer may still be incorrect if there's a flaw in the calculating methods. Now, whether the approximation is good enough in real life, thats a whole another thing.

The thing they teached us in general back in the day in poly that they most likely will be using methods to tell the truly clever ones out of the mere automatons .

Regards
Sam

[madmatt]

4 Ohms?

[awakefie]

I misread the title. At first I thought it said "Speaking of Psychics - quick homework question". Completely changes the meaning of the thread.

[ehque]

Originally Posted by Muaguana

Step 1. Look at equations.
Step 2. Solve for unknown variables.

Very simple, really. System of two equations with two unknowns. Plug and chug the numbers, and there's your answer.

Edit: Here's the composite equation to get the answer in one swing: T=(2*pi*r)/((G*M/r)^1/2), where T is the period, r is the distance from the asteroid to the sun (radius), G is the gravitational constant, and M is the mass of the sun. You might want to use SI units for distance and time, since the gravitational constant is in meters cubed per kilogram per seconds squared. Just take the answer in seconds you get and convert it to years (use the conversion factor 1 second = 3.16888 × 10^-8 year).

Given the nature of the question, I doubt he's supposed to solve it this way.

It's possible to solve the question without finding or knowing the mass of the sun, or G, since they factor into both equations and thus can be cancelled out.

I'm on a phone so the math is tricky, but i'll post 2 solutions here when i get to a real computer later today.

[ehque]

I'm back.

Kepler's 3rd Law: The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

You can memorise this or derive it from the above compound equations. Solving the problem should be easy now. If you are using G or M, you're overdoing the problem. If you are converting years to seconds, you're also overdoing it.

[Musiclogic]

Originally Posted by T-Bird

Hi.


The thing they teached us in general back in the day in poly that they most likely will be using methods to tell the truly clever ones out of the mere automatons .

Regards
Sam

Sam...did you say "Teached" LOL maybe Taught might adhere to grammar a bit better.

[Time Monkey]

Originally Posted by Muaguana

Step 1. Look at equations.
Step 2. Solve for unknown variables.

Very simple, really. System of two equations with two unknowns. Plug and chug the numbers, and there's your answer.

Edit: Here's the composite equation to get the answer in one swing: T=(2*pi*r)/((G*M/r)^1/2), where T is the period, r is the distance from the asteroid to the sun (radius), G is the gravitational constant, and M is the mass of the sun. You might want to use SI units for distance and time, since the gravitational constant is in meters cubed per kilogram per seconds squared. Just take the answer in seconds you get and convert it to years (use the conversion factor 1 second = 3.16888 × 10^-8 year).

heh?

[T-Bird]

Hi.

Originally Posted by Musiclogic

Sam...did you say "Teached" LOL maybe Taught might adhere to grammar a bit better.

True, thanks for pointing that out. I always try to use as good grammar as I can, but quite often I do fail .

Regards
Sam

[champbassist]

Originally Posted by ehque

Kepler's 3rd Law: The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

This is correct (i.e. gives the correct result), but the question mentions a circular orbit, not elliptical. Wonder why they'd say that

[guitar ed]

And the orbit is elliptical, not circular. In which case the answers is Null.

edg

[ehque]

A circle is a special form of ellipse, a form where the semi-major axis and the semi-minor axis are the same and known by one value - the radius. It's like how a square is a special form of rectangle.

Basically, Kepler's law is the general form and you can use a simplified version when you have circles.

EDIT: There are many ways of solving this question depending on how advanced the physics is.

The simplest is as mentioned above. Using Kepler's third law,

T_earth²r_asteroid³=T_asteroid²r_earth³

Substitute in values and solve for T_asteroid.

Alternatively, if you know the orbital period equation: where and hence a constant,

Deriving Kepler's third law from this equation is trivial, then you're back to solving a proportionality problem as above.

The question is most difficult when you have not memorised or do not know either Kepler's third law or the orbital period equation. In this case you will have to derive the equation from other equations like distance travelled/velocity, etc. It is even possible to flatten this problem into a 1-dimensional problem using SHM, if you are relatively desperate in an exam. You should still come to the same answer, but by a relatively roundabout route.

[MatticusMania]

42

[Funky Ghost]

It seems as if the simplest way to solve it is being asked for.

In order for gravity to be used you would need to know the masses interacting with the object in question. The nature of the orbits path ( angle of inclination ) would supply that information in a known solar system. Since that is not supplied it stands to reason it is not relevant to the problem.

Or am I missing something?

[ehque]

Originally Posted by Funky Ghost

It seems as if the simplest way to solve it is being asked for.

In order for gravity to be used you would need to know the masses interacting with the object in question. The nature of the orbits path ( angle of inclination ) would supply that information in a known solar system. Since that is not supplied it stands to reason it is not relevant to the problem.

Or am I missing something?

You can certainly derive GM (gravitational constant * solar mass) from the supplied data. It's supplied in a roundabout way, but you don't need it to solve the problem so there's no reason to derive it. The mass of earth and asteroid is irrelevant to the question, since orbital motion is independent from the mass of the orbiting object (as long as the mass is much smaller than the mass of the object being orbited).

The simplest way is already mentioned above. Use Kepler's third law and a simple proportionality equation.

[Funky Ghost]

Ok thank you.

Well, wait. Wouldn't all the masses have to be known, not just the solar mass? We use gravity wells as a way to slingshot earth objects.

I'm over thinking this aren't I?

* we both edited and you answered my question.

[champbassist]

Originally Posted by ehque

A circle is a special form of ellipse, a form where the semi-major axis and the semi-minor axis are the same and known by one value - the radius. It's like how a square is a special form of rectangle.

Oh yes. Didn't look at it that way.