Not really understanding Gravitational Potential energy or Kepler's laws?

[mayodt]

Hey, I have a test tommorow and I don't really understand gravitational potential energy or kepler's laws. It's not really a theory test, it's the math aspect, but I still don't understand it. I know of the formulas such as Fg=Gm1m2/r^2, Eg=Gm1m2/r, Ek=1/2 mv^2, and I think that's all. But somehow in the answer book for this question they come up with the equation C=GM/4(pie)^2 and C=r^3/T^2?

This is the question if it helps: Using the mass of the sun and the period of revolution of Venus around the Sun, determine the average Sun-Venus distance.

Also, I'm not really understanding a couple other questions as well, if anyone had something I could read that would help me out for this, it'd be amazing, thanks.

 

[berkeman]

Hey, I have a test tommorow and I don't really understand gravitational potential energy or kepler's laws. It's not really a theory test, it's the math aspect, but I still don't understand it. I know of the formulas such as Fg=Gm1m2/r^2, Eg=Gm1m2/r, Ek=1/2 mv^2, and I think that's all. But somehow in the answer book for this question they come up with the equation C=GM/4(pie)^2 and C=r^3/T^2?

This is the question if it helps: Using the mass of the sun and the period of revolution of Venus around the Sun, determine the average Sun-Venus distance.

Also, I'm not really understanding a couple other questions as well, if anyone had something I could read that would help me out for this, it'd be amazing, thanks.

I'm not much help on Kepler's Laws, but the wikipedia entry looks to have some good info:

http://en.wikipedia.org/wiki/Kepler's_laws

.

 

[Bloodthunder]

F = Gmm/r^2 for gravitational potential energy
F= mv/r = mr$\omega^{2}$for centripetal force.
$\omega$ = 2$\pi$/T

Reshuffle them.

 

[cupid.callin]

only third law of kepler has mathematical questions at basic level
$T^2 \propto R^3$
tell me if you need its derivation

for Gravitational potential ...

the GPE of 2 masses m1,m2 separated by r is $U = - \frac{Gm_1m_2}{r}$

for more help please write back ...

 

[rwisz]

Hi there,

Thank you for reaching out for help with understanding gravitational potential energy and Kepler's laws. These concepts can be quite complex, so it's understandable that you may be having trouble understanding them.

Gravitational potential energy is a measure of the potential energy stored in an object due to its position in a gravitational field. It is calculated using the formula Eg = Gm1m2/r, where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them. This formula tells us that as the distance between two objects increases, the gravitational potential energy decreases.

Kepler's laws, on the other hand, are three laws that describe the motion of planets around the sun. The first law, also known as the law of orbits, states that all planets move in elliptical orbits with the sun at one focus. The second law, also known as the law of areas, states that a line connecting a planet to the sun sweeps out equal areas in equal times. And the third law, also known as the law of periods, states that the square of the orbital period of a planet is directly proportional to the cube of the average distance from the sun.

To answer the question you provided, we can use the third law of Kepler to find the average Sun-Venus distance. The formula for this law is C = r^3/T^2, where C is a constant value for the entire solar system. We can rearrange this formula to solve for r, which gives us r = (C*T^2)^(1/3). To find the value of C, we can use the mass of the sun and the period of revolution of Venus. The mass of the sun is approximately 1.989 x 10^30 kg, and the period of revolution of Venus is approximately 224.7 Earth days, or 0.615 Earth years. Plugging these values into the formula C = GM/4(pi)^2, we get C = (6.674 x 10^-11 * 1.989 x 10^30)/4(pi)^2 = 1.327 x 10^20 m^3/s^2. Now, we can plug this value of C and the period of revolution of Venus into our formula for r, giving us r = (1.327 x 10^20 * 0.615^2)^(1/3) =