# Calculating Kinetic & Potential Energies of Earth at Perihelion/Aphelion

• frisky
In summary, you calculate the kinetic and potential energies of the Earth at the perihelion and aphelion by using the following equation: v=√(GM/R), where M is the sun's mass, R is the Earth's orbital radius, and GM is the total mass of the sun and Earth. The kinetic energy, E, is 1/2GMm/R, and the Earth's potential energy, U, is -GMm/R. However, the change in E between the perihelion and aphelion is (1/2)GMm/[(1/R) - (1/r)] where R and r are the respective orbital radii. The change in U between
frisky
How do you calculate the kinetic and potential energies of the Earth at the perihelion and aphelion?

Given that the Earth's orbit is nearly circular

GMm/R^2 = m(v^2)/R where M = sun's mass, m = Earth's R = orbital radius

so

v = √(GM/R)

The kinetic energy, E, would be

E = (1/2)m(v^2) = (1/2)GMm/R

The Earth's potential energy, U, would be

U = -GMm/R

However, the change in E between the perihelion and aphelion would be

(1/2)GMm/[(1/R) - (1/r)] where R and r are the respective orbital radii;

while the change in U would be

-GMm/[(1/R) - (1/r)] .

Given conservation of energy, the last two terms can not both be right.

This brings me back to the original question.

frisky said:
How do you calculate the kinetic and potential energies of the Earth at the perihelion and aphelion?

Given that the Earth's orbit is nearly circular

GMm/R^2 = m(v^2)/R where M = sun's mass, m = Earth's R = orbital radius

so

v = √(GM/R)

The kinetic energy, E, would be

E = (1/2)m(v^2) = (1/2)GMm/R

The Earth's potential energy, U, would be

U = -GMm/R

However, the change in E between the perihelion and aphelion would be

(1/2)GMm/[(1/R) - (1/r)] where R and r are the respective orbital radii;
This is not correct. What you have calculated here is the difference in E between two circular orbits of radii r and R. With a ellipitical orbit the orbital velocity at perhelion is greater than the circular orbital velocity for that radius and the orbital velocity at aphelion is less than the circular orbital velocity for that radius.
while the change in U would be

-GMm/[(1/R) - (1/r)] .

Given conservation of energy, the last two terms can not both be right.

This brings me back to the original question.

Let's see if I can give you a push in the right direction:

The equations:

$$V = \sqrt{\frac{GM}{r}}$$
and
$$E = \frac{GMm}{2r}$$

Are only true for an elipitical orbit when r = A (The semimajor axis of the orbit, or average orbital distance).

$$U = -\frac{GMm}{r}$$

is true for all points of the orbit.

So what is the total energy (E+U) at a radius A?
is This the same as the total energy at r and R?

Hopefully, this will give you enough to go from here.

You're using rough estimates (treating the orbit as circular for each velocity, for example), so it won't cancel out. If you use exact values, it will.

First off, it's easier if you use the specific energy per unit of mass to analyze the motion, then multiply the mass back in if you need an actual value for energy.
$$E=m\epsilon$$ where $$\epsilon$$ is specific energy
$$m\epsilon=m(\frac{1}{2}v^2-\frac{GM}{r})$$
$$\epsilon=\frac{1}{2}v^2-\frac{GM}{r}$$

If you rearrange this:

$$v=\sqrt{2\epsilon+\frac{2GM}{r}$$

E is also equal to:

$$\epsilon=-\frac{GM}{a}$$

Substituting and rearranging, you get Leibniz's vis-viva equation:

$$v=\sqrt{GM(\frac{2}{r}-\frac{1}{a})}$$
(If you have a circular orbit, the radius and semi-major axis are equal, which reduces to the equation you used.) If you use this equation for velocity, your calculations will show energy is conserved. (substitute the velocity equation into the specific kinetic energy equation)

You have to calculate the specific potential energy for both perihelion and apohelion, as well. In other words, substitute the radius of perihelion for R and the radius of apohelion for R.

If you need the actual energy of the orbit, multiply the mass back in, but it's not necessary to show energy is conserved.

Thanx.

I understand this now.

## What is the definition of kinetic energy and potential energy?

Kinetic energy is the energy an object possesses due to its motion. It is directly proportional to an object's mass and the square of its velocity. Potential energy is the energy an object possesses due to its position or configuration. It can be gravitational, chemical, or elastic in nature.

## How do you calculate the kinetic energy of Earth at perihelion/aphelion?

To calculate the kinetic energy of Earth at perihelion/aphelion, you need to know the mass of the Earth, its velocity at the specific point, and the gravitational constant. The formula is KE = (1/2)mv^2, where m is the mass of the Earth and v is its velocity.

## How do you calculate the potential energy of Earth at perihelion/aphelion?

To calculate the potential energy of Earth at perihelion/aphelion, you need to know the mass of the Earth, the distance between the Earth and the Sun at the specific point, and the gravitational constant. The formula is PE = -GmM/r, where m is the mass of the Earth, M is the mass of the Sun, r is the distance between them, and G is the gravitational constant.

## How does the kinetic and potential energy of Earth change at perihelion/aphelion?

At perihelion, the Earth is closest to the Sun and therefore has a higher kinetic energy due to its increased velocity. However, its potential energy is lower as the distance between the Earth and the Sun is smaller. At aphelion, the Earth is farthest from the Sun and has a lower kinetic energy due to its decreased velocity, but its potential energy is higher due to the increased distance between the Earth and the Sun.

## What are the implications of calculating the kinetic and potential energy of Earth at perihelion/aphelion?

Calculating the kinetic and potential energy of Earth at perihelion/aphelion helps us understand the dynamics of the Earth's orbit around the Sun and its changing speed and distance. It also allows us to predict future changes in the Earth's orbit and its impact on the Earth's climate and seasons. Additionally, it provides valuable information for space missions and spacecraft navigation around the Earth.

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