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 = T_D * [\frac{a_R}{a_D} ]^{\frac{3}{2} }](https://tex.z-dn.net/?f=T_R%20%20%3D%20T_D%20%2A%20%20%5B%5Cfrac%7Ba_R%7D%7Ba_D%7D%20%5D%5E%7B%5Cfrac%7B3%7D%7B2%7D%20%7D)
=> ![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:
F=m(11.8m/s²)
For example, if m=10,000kg, F=118,000N.
Explanation:
There are only two vertical forces acting on the rocket: the force applied from its thrusters F, and its weight mg. So, we can write the equation of motion of the rocket as:
Solving for the force F, we obtain that:
Since we know the values for a (2m/s²) and g (9.8m/s²), we have that:
From this relationship, we can calculate some possible values for F and m. For example, if m=10,000kg, we can obtain F:
In this case, the force from the rocket's thrusters is equal to 118,000N.
Explanation:
the acceleration will be unchanged according to newton second law of motion
Acceleration is measured in meters per second square.
