lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A in a circular orbit of radius 5000 km and period 4.0 hours. What is the radius of Planet R47A

Explanation:

The acceleration due to gravity g is defined as

and solving for R, we find that

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration experienced by the satellite is equal to the gravitational force or

The orbital velocity v is the velocity of the satellite around the planet defined as

where r is the radius of the satellite's orbit in meters and T is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

Solving for M, we get

Putting this expression back into Eqn(1), we get

a

explanation:

no diagram was provided, but i'm guessing it's the one i've attached to my answer.

here, a mechanical force turns the rod.   this generates an electric current in the coil, which powers the bulb.

i think it's 35 meters