Energy of Earth Orbiting Around the Sun

[Ishaan S]

Homework Statement

The Earth's distance from the sun varies from 1.471 x 10⁸ km to 1.521 x 10⁸ km during the year. Determine the difference in (a) the potential energy, (b) the earth's kinetic energy, and (c) the total energy between these extreme points. Take the sun to be at rest.

Homework Equations

F_Grav = (GMm)/r²

U = integral(a→b)(GMm/r²)dr

KE = .5mv²

Law of the Conservation of energy: E_i = E_f

a_c = v² / r

The Attempt at a Solution

I got (a) easily. I need help with b and c, though.

I tried taking the average distance from the sun, which is just the length of the semi-major axis, which is

1.521 x 10⁸ km - 1.471 x 10⁸ km = 5.00 x 10⁹ m = r_ave

Then I used the centripetal acceleration formula to get

F_a = F_G = (GM_sun / (r_ave)²) = (v²)_ave/r_ave.

Solving for v_ave I get

v_ave = 2.7 x 10¹⁰ m/s.

I then calculated average kinetic energy to be

.5 * m_earth * (v_ave)² = 2.1 * 10⁴⁵ J.
Average gravitational energy would be

-GMm/r_ave = -1.6 * 10³⁵ J.

Average total energy would be average kinetic + average potential.

Since energy is constant the average energy is constant for instantaneous energy.

Therefore, at perihelion and aphelion, I would have average total energy - potential energy at the point, correct?

Any help would be appreciated, thanks.

 

[SteamKing]

Homework Statement

The Earth's distance from the sun varies from 1.471 x 10⁸ km to 1.521 x 10⁸ km during the year. Determine the difference in (a) the potential energy, (b) the earth's kinetic energy, and (c) the total energy between these extreme points. Take the sun to be at rest.

Homework Equations

F_Grav = (GMm)/r²

U = integral(a→b)(GMm/r²)dr

KE = .5mv²

Law of the Conservation of energy: E_i = E_f

a_c = v² / r

The Attempt at a Solution

I got (a) easily. I need help with b and c, though.

I tried taking the average distance from the sun, which is just the length of the semi-major axis, which is

1.521 x 10⁸ km - 1.471 x 10⁸ km = 5.00 x 10⁹ m = r_ave

This calculation makes no sense. How can you subtract two positive numbers and have the difference be greater than either number?

Also, it's not even how you calculate an average distance.

Then I used the centripetal acceleration formula to get

F_a = F_G = (GM_sun / (r_ave)²) = (v²)_ave/r_ave.

Solving for v_ave I get

v_ave = 2.7 x 10¹⁰ m/s.

Really? What's the speed of light in a vacuum? Do you have the Earth going faster or slower than light itself?

http://en.wikipedia.org/wiki/Speed_of_light

I then calculated average kinetic energy to be

.5 * m_earth * (v_ave)² = 2.1 * 10⁴⁵ J.
Average gravitational energy would be

-GMm/r_ave = -1.6 * 10³⁵ J.

Average total energy would be average kinetic + average potential.

Since energy is constant the average energy is constant for instantaneous energy.

Therefore, at perihelion and aphelion, I would have average total energy - potential energy at the point, correct?

Any help would be appreciated, thanks.

Correct your mistakes as pointed out above.

 

[Ishaan S]

OK.

The mean distance is (1.521*10⁸ km + 1.471*10⁸ km)/2 = 1.496*10⁸ km = 1.496*10¹¹ m.

Then the average velocity comes from the centripetal acceleration formula:

F_a = F_G = GM/(r_ave)² = (v_ave)² / r_ave

Solve for v_ave

v_ave = 30,000 m/s.

So average KE = ,5mv_ave² = 2.7*10³³ J.

Average potential energy is

-GMm/r_ave = -5.3 * 10³³ J.

So total energy is U + KE = -2.7*10³³ J.

So at perihelion the KE would be

E_total - (-GMm/r_{closest distance}) = 2.8*10³³ J.

At aphelion the KE would be

E_total - (-GMm/r_{far distance}) = 2.6*10³³ J.

Is this correct?

 

[haruspex]

I do not understand how an 'average' distance could be helpful in answering the question. Just find the PE and KE at the given extremes.
Further, it is not clear what would be meant by an average distance from the sun. Taking the average of the two extreme distances is unlikely to produce anything useful. An integral would make more sense, but even then there are choices to be made. You could take an average over time, or an average over theta, etc.