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:
1. 37.8J
2. 18 Billion Joules, 18 Gigajoules
3. 9.81 Billion Joules, 9.81 Gigajoules
Explanation:
Use the formulas provided,
KE=(1/2)mv^2 and PE=mgh, noting that g=9.81
Answer:
b Day-to-day condition of the atmosphere
Explanation:
Weather is short term, Climate is long term
Answer:
A) -2N
B) Left
C) -0.5
Explanation:
A) -12 + 10
B) More force is acted on in that direction
C) Net force/Mass (-2/4)
Answer:
m = 788.2[kg]
Explanation:
The potential energy of a body is defined as the product of mass by gravitational acceleration by height. And it can be calculated by means of the following equation.

where:
Epot = potential energy = 63405 [J]
m = mass [kg]
g = gravity acceleration = 9.81[m/s²]
h = elevation = 8.2[m]
Now replacing:
![63405=m*9.81*8.2\\m=788.2[kg]](https://tex.z-dn.net/?f=63405%3Dm%2A9.81%2A8.2%5C%5Cm%3D788.2%5Bkg%5D)