Answer:
Explanation:
We can solve the problem by using Kepler's third law, which states that the ratio between the cube of the orbital radius and the square of the orbital period is constant for every object orbiting the Sun. So we can write
where
is the distance of the new object from the sun (orbital radius)
is the orbital period of the object
is the orbital radius of the Earth
is the orbital period the Earth
Solving the equation for , we find
![r_o = \sqrt[3]{\frac{r_e^3}{T_e^2}T_o^2} =\sqrt[3]{\frac{(1.50\cdot 10^{11}m)^3}{(365 d)^2}(180 d)^2}=9.4\cdot 10^{10} m](https://tex.z-dn.net/?f=r_o%20%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7Br_e%5E3%7D%7BT_e%5E2%7DT_o%5E2%7D%20%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B%281.50%5Ccdot%2010%5E%7B11%7Dm%29%5E3%7D%7B%28365%20d%29%5E2%7D%28180%20d%29%5E2%7D%3D9.4%5Ccdot%2010%5E%7B10%7D%20m)
Answer:
See Below
Explanation:
Okay, I thinkkk what it is asking by what you summarzied for me issss:
They split the total time into four quarters. They then took (for the first quarter) the start time. Then when the first quarter ends and the second quarter starts is the "end" time.
They then subtract the start time of the second quarter from the end time of the first quarter.
I hope this helps, good luck! :D
It increases. Mercury takes 88 days to orbit the sun once. The Earth takes a year. Pluto takes 248 years.
Pretty sure it’s A. Hope this helps.
Answer:
D. 2^(3/2)
Explanation:
Given that
T² = A³
Let the mean distance between the sun and planet Y be x
Therefore,
T(Y)² = x³
T(Y) = x^(3/2)
Let the mean distance between the sun and planet X be x/2
Therefore,
T(Y)² = (x/2)³
T(Y) = (x/2)^(3/2)
The factor of increase from planet X to planet Y is:
T(Y) / T(X) = x^(3/2) / (x/2)^(3/2)
T(Y) / T(X) = (2)^(3/2)
