Answer on Question #61300 - Physics - Mechanics | Relativity
Question:
1. IF a small planet were discovered whose orbital period was twice that of Earth, how many time
farther
from
the
Sun
would
this
planet
be?
2. Determine Kepler's constant for any satellite of Earth.
Solution:
1.
The third Kepler’s law is:
𝑇12
=
𝑇22
𝑎13
𝑎23
𝑇12 𝑎13
𝑎1 3 𝑇12
=
⇒
= √ 2,
𝑎2
𝑇22 𝑎23
𝑇2
where 𝑎2 is the radius of Earth's orbit and 𝑎1 is the radius of planet's orbit, 𝑇2 and 𝑇1 are orbital
periods of Earth and planet.
1
𝑎1 3 𝑇12 3 (2𝑇2 )2
= √ 2 = √ 2 = 43 ≈ 1.5874 …
𝑎2
𝑇2
𝑇2
2.
Kepler’s constant is: 𝐾 =
For Earth: 𝐾 =
𝐺𝑀
4𝜋2
=
𝑎3
𝑇2
=
𝐺𝑀
4𝜋2
6.67∙10−11 ∙6∙1024
4𝜋2
𝑚3
≈ 1.014 ∙ 1013 (
𝑠2
)
Answer:
1)
𝑎1
𝑎2
≈ 1.5874 …
2) 𝐾 ≈ 1.014 ∙ 1013
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