Finding distance of a planet whose orbital period is 2x Earth

In summary, according to Kepler's 3rd law of planetary motion, the orbital radius of a small planet that has an orbital period twice that of Earth can be calculated using the equation rp = 3√([ (2 x TE)2 / TE2 ] rE3). However, if the orbital period is doubled, the radius of the planet should be √2x instead of 2x. It may be helpful to re-arrange the equation to eliminate a variable and reduce the chances for math errors.
  • #1
JohnGalt
1
0
A small planet has an orbital period that is twice that of Earth. What is the planet's orbital distance?

From Kepler's 3rd law of planetary motion, I can assume that:

rp3 / rE3 = Tp2 / TE2

where rp is the orbital radius of the planet and Tp is the orbital period of the planet

Therefore,

rp3 = (Tp2 / TE2) rE3

= [ (2 x TE)2 / TE2 ] rE3

Therefore,

rp = 3√ ([ (2 x TE)2 / TE2 ] rE3)

= 1 060 071 250 m

which differs from the given answer. What error have I made?
 
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  • #2
I imagine this is a shortcut but, if the period is 2x, isn't the radius simply √2x?

(No, it's other way round. If the radius is 2x, then the orbital velocity is √2x.)
 
  • #3
The equation looks okay, did you use the right values?

Also, reducing the equation further might help as it reduces chances for math errors.

(hint: by some slight re-arranging, you can eliminate a variable from the equation.)
 

1. How is the distance of a planet with a 2x Earth's orbital period calculated?

The distance of a planet with a 2x Earth's orbital period is calculated using Kepler's Third Law, which states that the square of a planet's orbital period is directly proportional to the cube of its semi-major axis (distance from the sun). The equation for this is: a^3 = (P^2 * GM) / 4π^2, where a is the semi-major axis, P is the orbital period, and GM is the gravitational constant multiplied by the mass of the sun.

2. What is the semi-major axis of a planet with a 2x Earth's orbital period?

The semi-major axis of a planet with a 2x Earth's orbital period can be calculated using the equation mentioned above: a = (P^2 * GM)^(1/3) / 4π^2. The result will be in units of astronomical units (AU), which is the average distance between the Earth and the sun.

3. How does the distance of a planet with a 2x Earth's orbital period compare to Earth's distance from the sun?

The distance of a planet with a 2x Earth's orbital period will be approximately 1.58 times greater than Earth's distance from the sun. This is because the semi-major axis is directly proportional to the orbital period, and in this case, the orbital period is doubled.

4. Can the distance of a planet with a 2x Earth's orbital period change?

Yes, the distance of a planet with a 2x Earth's orbital period can change over time due to various factors such as gravitational interactions with other planets, tidal forces, and even changes in the planet's own mass. However, these changes are usually small and occur over long periods of time.

5. How is the distance of a planet with a 2x Earth's orbital period measured or observed?

The distance of a planet with a 2x Earth's orbital period can be measured or observed using various methods, such as radar ranging, parallax measurements, and astrometry. These techniques involve using instruments like telescopes and mathematical equations to calculate the planet's distance from the sun. The data collected from these methods can also be used to refine the planetary orbit and determine any changes in distance over time.

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