By using the orbital period equation we will find that the orbital radius is r = 4.29*10^11 m
<h3>
What is the orbital period?</h3>
This would be the time that a given body does a complete revolution in its orbit.
It can be written as:
Where:
- π = 3.14
- G is the gravitational constant = 6.67*10^(-11) m^3/(kg*s^2)
- M is the mass of the sun = 1.989*10^30 kg
- r is the radius, which we want to find.
Rewriting the equation for the radius we get:
![T = \sqrt{\frac{4*\pi ^2*r^3}{G*M} }\\\\r = \sqrt[3]{ \frac{T^2*G*M}{4*\pi ^2} }](https://tex.z-dn.net/?f=T%20%3D%20%5Csqrt%7B%5Cfrac%7B4%2A%5Cpi%20%5E2%2Ar%5E3%7D%7BG%2AM%7D%20%7D%5C%5C%5C%5Cr%20%3D%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7BT%5E2%2AG%2AM%7D%7B4%2A%5Cpi%20%5E2%7D%20%7D)
Where T = 7.5 years = 7.5*(3.154*10^7 s) = 2.3655*10^8 s
Replacing the values in the equation we get:
![r = \sqrt[3]{ \frac{(2.3655*10^8 s)^2*(6.67*10^{-11} m^3/(kg*s^2))*(1.989*10^{30} kg)}{4*3.14 ^2} } = 4.29*10^{11 }m](https://tex.z-dn.net/?f=r%20%3D%20%5Csqrt%5B3%5D%7B%20%5Cfrac%7B%282.3655%2A10%5E8%20s%29%5E2%2A%286.67%2A10%5E%7B-11%7D%20m%5E3%2F%28kg%2As%5E2%29%29%2A%281.989%2A10%5E%7B30%7D%20kg%29%7D%7B4%2A3.14%20%5E2%7D%20%7D%20%3D%204.29%2A10%5E%7B11%20%7Dm)
So the orbital radius is 4.29*10^11 m
If you want to learn more about orbits, you can read:
brainly.com/question/11996385
Yes because it helps with my work
<h2>
Answer:</h2>
The earth behaves as a magnetic dipole. Therefore a freely suspended magnet always points towards in the north-south direction because the north pole of the suspended magnet attracts the south pole of the earth's magnet which is the geographical north pole of the earth.
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Answer:
Explanation:
From the Question we are told that:
Mass 
Coefficient of kinetic friction 
Generally the equation for Frictional force is mathematically given by
Generally the Newton's equation for Acceleration due to Friction force is mathematically given by
Therefore
