A 21.0-kg satellite has a circular orbit with a period of 2.37 h and a radius of 8.50×106 m around a planet of unknown mass. If the magnitude of the gravitational acceleration on the surface of the planet is 7.10 m/s2, what is the radius of the planet?

Use Kepler's third law (with period and orbit radius) to get the Mass of the planet.

The gravitation acceleration at the planet's surface is g' = GM/R^2.
Since you now know g' and M, you can solve for R. (G is the universal gravitational constant.)

great! thanks again for your help :)

You're welcome! Let's work through the problem step by step.

First, let's calculate the mass of the planet using Kepler's third law. Kepler's third law states that the square of the period of an orbit is proportional to the cube of the semi-major axis of the orbit.

T^2 = (4π^2/GM) * R^3

Where T is the period, G is the universal gravitational constant, M is the mass of the planet, and R is the radius of the orbit.

Given that T = 2.37 hours and R = 8.50×10^6 m, we want to solve for M.

We can rearrange the equation to solve for M:

M = (4π^2/G) * (R^3 / T^2)

Now we need to know the value of the universal gravitational constant, G, which is approximately 6.67430 × 10^-11 m^3/(kg·s^2).

Substituting the known values into the equation, we have:

M = (4π^2 / (6.67430 × 10^-11)) * ((8.50×10^6)^3 / (2.37^2))

Evaluating this expression will give us the mass of the planet.

Once we have the mass of the planet, we can calculate the radius, R. We know that the gravitational acceleration at the planet's surface is given as 7.10 m/s^2, which we can denote as g'. We can also express g' as GM/R^2, where M is the mass of the planet and R is the radius of the planet.

7.10 m/s^2 = (GM) / R^2

We already know the mass of the planet (M) from the earlier calculation. Rearranging the equation, we can solve for R:

R = √((GM) / g')

Substituting the calculated values for M and g', we can find the radius of the planet.

I hope this explanation helps! Let me know if you have any further questions.