Calculating Earth's Orbital Velocity from Varying Distance to the Sun

[vela]

Perfect!

 

[toothpaste666]

so

$V_n = (2(\frac{V_n R_n}{R_f} + G\frac{M_s}{R_f} - G\frac{M_s}{R_n})^\frac{1}{2}))$

?

 

[toothpaste666]

wait no hold on i forgot the rest of the KE equation

 

[toothpaste666]

$V_n = (2(\frac{ M_e V_n^2 R_n^2}{2 R_f^2} + G\frac{M_s}{R_f} - G\frac{M_s}{R_n})^\frac{1}{2}))$

 

[vela]

The mass of the Earth $M_e$ shouldn't be there.

 

[toothpaste666]

ahh right those canceled. Sorry that was me being sloppy.

$V_n = (2(\frac{V_n^2 R_n^2}{2 R_f^2} + G\frac{M_s}{R_f} - G\frac{M_s}{R_n})^\frac{1}{2}))$

and then after plugging in i can go back and solve for

$V_f = \frac{V_n R_n}{R_f}$

 

[vela]

You still have sign errors. Remember that gravitational potential energy is negative.

 

[toothpaste666]

So if i flipped the signs on the potential energy i end up with this:

$V_n = (2(\frac{V_n^2 R_n^2}{2 R_f^2} - G\frac{M_s}{R_f} + G\frac{M_s}{R_n})^\frac{1}{2}))$

would you be able to explain why its negative a little more? I've done other problems with energy like the ones where a rollercoaster goes from the top of a hill to the bottom and you have to find the velocity at the bottom, and in those I used mgy as the potential energy, i still think that's gravitational energy but in those problems it was fine to write it as positive. How can i tell when to change the sign?

Also what process could i have used to use F = MA to rewrite Vf instead of the circular sector approach?

 

[ehild]

You took the potential energy as GM_nM_f/R. It is zero at infinity, isn't it?

If the Earth gets closer to the Sun, its potential energy should decrease, just as the PE of falling stone on the Earth. And the kinetic energy increases, but the KE is always positive.

If a function decreases from zero, it gets negative. When R decreases the magnitude of the potential energy increases as you divide by R. So there should be a minus sign so as the PE decrease: PE=-GM_nM_f/R.

You can choose, where is the zero of the potential energy. Near to the surface of Earth it is convenient to chose it zero on the ground. At the top of a hill it is positive then, but decreases when the stone falls down. You can choose the top of hill where the potential energy is zero. Then its gets negative when the stone falls.

ehild

 

[toothpaste666]

ahh ok thanks I think I get it. Aren't there parts of the orbit where potential energy is increasing as the Earth moves farther away from the sun?

 

[ehild]

If the Earth goes farther away from the Sun, its potential energy increases, which means, it becomes less negative. And if it goes very -very far, its potential energy will approximate zero.


ehild

 

[vela]

Also what process could i have used to use F = MA to rewrite Vf instead of the circular sector approach?

You have to recognize that the Earth's orbit is essentially a circle because the eccentricity of the orbit is nearly 0. The Earth's centripetal acceleration is given by $v^2/\rho$, where $\rho$ is the radius of Earth's orbit, which you can express in terms of $R_f$ and $R_n$.

 

[toothpaste666]

thank you both so much!