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

[toothpaste666]

Homework Statement

Given that the Earth's distance from the sun varies from 1.47 to 1.52x10^11m, determine the minimum and maximum velocities of the Earth in it's orbit around the sun.

Homework Equations

F=G\frac{m1m2}{r^2}

E=K+U ?

The Attempt at a Solution

I think the way to do this is with K1+U1 = K2+U2 , where one side of the equation is the Earth at its closest point to the sun and the other side is the Earth at its farthest point. Let Me = mass of earth, Ms = mass of sun, Rn = distance at nearest point, Rf= distance at farthest point, Vn = velocity at nearest point, Vf = velocity at farthest point.

K1+U1 = K2 + U2

\frac{MeVn^2}{2} + G\frac{MsMe}{Rn} =\frac{MeVf^2}{2} + G\frac{MsMe}{Rf}

the Me's cancel. to solve for Vn replace Vf with \frac{2piRf}{T}

\frac{Vn^2}{2} + G\frac{Ms}{Rn} =\frac{2pi^2Rf^2}{T^2} + G\frac{Ms}{Rf}

\frac{Vn^2}{2} =\frac{2pi^2Rf^2}{T^2} + G\frac{Ms}{Rf} - G\frac{Ms}{Rn}


Vn^2 = 2(\frac{2pi^2Rf^2}{T^2} + G\frac{Ms}{Rf} - G\frac{Ms}{Rn})

Vn = (2(\frac{2pi^2Rf^2}{T^2} + G\frac{Ms}{Rf} - G\frac{Ms}{Rn}))^\frac{1}{2}

then after plugging in I would go back and solve for Vf. Would this give me the correct answer?

 

[BiGyElLoWhAt]

well, one thing I want to say is that the Earth is revolving, not translating. You don't want 1/2 mv^2, you want 1/2 Iw^2 for your KE, and I'm sure Mr^2 is good enough for I. other than that I think you're on the right track.

 

[BiGyElLoWhAt]

... I was just looking at your eq. again, and I'm not sure what you're doing with the 2pi*r/T, that is w, but I don't think you're looking at the KE in this situation correctly.

 

[toothpaste666]

Ok cool thanks. What does Iw represent?

 

[toothpaste666]

I was using the equations for circular motion where v = 2(pi)r/T . I am only vaguely sure of what I am doing though.

 

[BiGyElLoWhAt]

ok, looking again (again)... that'll get you the tangential velocity, but not angular, I'm not sure what you're after exactly.

Also, I think it's good practice to start with the angular and convert over, seeing as you have KE as a result of rotation, not translation. I think that will get you the correct result, though (this time).

 

[BiGyElLoWhAt]

Iw is moment of inertia * angular velocity

 

[toothpaste666]

what exactly is the angular velocity? I am not sure If I am up to that yet in my physics class. The section this problem was from was covering universal gravitation as well as kepler's laws. The question just says find the maximum and minimum velocities it doesn't specify between angular or tangential. The KE I am using won't work?

 

[BiGyElLoWhAt]

well angular velocity is the rate of revolution measured in radians/seconds, or also commonly 1/seconds (radians are funny that way). w = 2pi/T =v/r, so what you're doing will work I think.

 

[BiGyElLoWhAt]

just out of curiosity is it a general physics course?

 

[Chestermiller]

Suppose it was a perfectly circular orbit around the sun, at the average distance of 1.495x10^11 m. Would you know how to determine the tangential velocity then (say using F = ma)? What value would you get for the tangential velocity?

Chet

 

[vela]

Homework Statement

Given that the Earth's distance from the sun varies from 1.47 to 1.52x10^11m, determine the minimum and maximum velocities of the Earth in it's orbit around the sun.

Homework Equations

F=G\frac{m1m2}{r^2}

E=K+U ?

The Attempt at a Solution

I think the way to do this is with K1+U1 = K2+U2 , where one side of the equation is the Earth at its closest point to the sun and the other side is the Earth at its farthest point. Let Me = mass of earth, Ms = mass of sun, Rn = distance at nearest point, Rf= distance at farthest point, Vn = velocity at nearest point, Vf = velocity at farthest point.

K1+U1 = K2 + U2

\frac{MeVn^2}{2} + G\frac{MsMe}{Rn} =\frac{MeVf^2}{2} + G\frac{MsMe}{Rf}

the Me's cancel. to solve for Vn replace Vf with \frac{2piRf}{T}

\frac{Vn^2}{2} + G\frac{Ms}{Rn} =\frac{2pi^2Rf^2}{T^2} + G\frac{Ms}{Rf}

\frac{Vn^2}{2} =\frac{2pi^2Rf^2}{T^2} + G\frac{Ms}{Rf} - G\frac{Ms}{Rn}Vn^2 = 2(\frac{2pi^2Rf^2}{T^2} + G\frac{Ms}{Rf} - G\frac{Ms}{Rn})

Vn = (2(\frac{2pi^2Rf^2}{T^2} + G\frac{Ms}{Rf} - G\frac{Ms}{Rn}))^\frac{1}{2}

then after plugging in I would go back and solve for Vf. Would this give me the correct answer?

Using $V_f = \frac{2\pi R_f}{T}$ isn't correct. This assumes that the Earth covers a distance of $2\pi R_f$ in one year, which would be true if it followed a circular path of radius $R_f$, but the Earth covers less distance because it moves toward the Sun.

You have two unknowns, so you're going to need another equation. You used conservation of energy already (though the sign of the potential energy is wrong). What's another conserved quantity you can use?

 

[vela]

well, one thing I want to say is that the Earth is revolving, not translating. You don't want 1/2 mv^2, you want 1/2 Iw^2 for your KE, and I'm sure Mr^2 is good enough for I. other than that I think you're on the right track.

This is completely wrong. You've misread the problem.

 

[toothpaste666]

Using $V_f = \frac{2\pi R_f}{T}$ isn't correct. This assumes that the Earth covers a distance of $2\pi R_f$ in one year, which would be true if it followed a circular path of radius $R_f$, but the Earth covers less distance because it moves toward the Sun.

You have two unknowns, so you're going to need another equation. You used conservation of energy already (though the sign of the potential energy is wrong). What's another conserved quantity you can use?

I think energy is the only conserved quantity I know so far. The only one that definitely stays the same over time. Also, do you mean another equation involving velocity?

 

[ehild]

What about angular momentum?

ehild

 

[vela]

It's a central force, so angular momentum is conserved.

 

[toothpaste666]

I haven't learned about momentum yet. There is no other way to solve this?

 

[vela]

Use F=ma to get another relationship between V and R.

 

[ehild]

I haven't learned about momentum yet. There is no other way to solve this?

You certainly know Kepler's Laws. What do the first two say?

ehild

 

[toothpaste666]

do i have to use T^2 = r^3

 

[toothpaste666]

no wait that's the third. the first one says that the orbits of planets around the sun are ellipses. The second law says that each planets moves so that a line from the planet to the sun sweeps out equal areas in equal time

 

[ehild]

no wait that's the third. the first one says that the orbits of planets around the sun are ellipses. The second law says that each planets moves so that a line from the planet to the sun sweeps out equal areas in equal time

That is it, the second one. What are the areas the line from the planet to the Sun sweeps in one second when the planet is nearest and when it is farthest? For that short time you can consider those parts of the ellipse as they were sectors of circles. How do you calculate the area of a circular sector?

ehild

 

[toothpaste666]

the area would be (theta * r^2)/2 and since it is going from the farthest point to the nearest point its only sweeping out half the ellipse so theta would be pi?

 

[vela]

I think using F=ma would be simpler.

 

[ehild]

the area would be (theta * r^2)/2 and since it is going from the farthest point to the nearest point its only sweeping out half the ellipse so theta would be pi?

I mean only thin sectors, swept by the radius in 1 second at the nearest and farthest points.
You know that the area of a circular sector is sr/2, where s is the length of arc and r is the radius of the circle. Supposing the speed of the planet is v, what is the length s it covers in 1 s?

ehild

 

[toothpaste666]

I'm sorry guys Idk why I'm having so much trouble with this. Can I use V=D/T where D = s (the distance swept out) and V = the velocity of Earth at whichever of the two points you are looking at. T = 1 so V will just be equal to s ?

 

[ehild]

I'm sorry guys Idk why I'm having so much trouble with this. Can I use V=D/T where D = s (the distance swept out) and V = the velocity of Earth at whichever of the two points you are looking at. T = 1 so V will just be equal to s ?

Yes, s=vt and t=1. The areas are equal. What relation follows from that between the speeds and distances from Sun?

ehild

 

[toothpaste666]

that the velocities are also equal?

 

[vela]

Don't just guess. Work it out based on what ehild said.

 

[toothpaste666]

ok so the area of one sector is

\frac{s_1 R_n}{2}

the other is

\frac{s_2 R_f}{2}

the areas are equal so

\frac{s_1 R_n}{2} = \frac{s_2 R_f}{2}

s_1 R_n = s_2 R_f

s1 = Vn and s2 = Vf

V_n R_n = V_f R_f

V_f = \frac{V_n R_n}{R_f}

 

[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!