Complete Question
Planet D has a semi-major axis = 60 AU and an orbital period of 18.164 days. A piece of rocky debris in space has a semi major axis of 45.0 AU. What is its orbital period?
Answer:
The value is
Explanation:
From the question we are told that
The semi - major axis of the rocky debris 
The semi - major axis of Planet D is 
The orbital period of planet D is 
Generally from Kepler third law

Here T is the orbital period while a is the semi major axis
So

=>
=> ![T_R = 18.164 * [\frac{ 45}{60} ]^{\frac{3}{2} }](https://tex.z-dn.net/?f=T_R%20%20%3D%2018.164%20%20%2A%20%20%5B%5Cfrac%7B%2045%7D%7B60%7D%20%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D)
=>
Answer:
Ф = 239.73 rad
Explanation:
α = 12 + 15×t
W = ∫α×dt
= ∫(12 + 5×t)×dt
= 12×t + 2.5×t^2
then:
Ф = ∫W×dt
= ∫(12×t + 2.5×t^2)dt
= 6×t^2 + 5/6×t^3
therefore the angle at t = 4.88s is:
Ф = 6×(4.88)^2 + 5/6×(4.88)^3
= 239.73 rad
Answer:
d. Gravity
Explanation:
In the beginning of their lives, stars are formed when clouds of dust and gas are pulled together by gravity.
Hope this helps!!
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