A planet with mass 7 x 1024 kg orbits a sun with a period of 1 Earth's years. The speed of the planet around that sun is constant and |v|=43000 m/s.

What is the mass of that sun?

The period of an orbiting body derives from

....T = 2(Pi)sqrt(r^3/µ) in seconds where T = the period in sec, Pi = 3.1416, r = the orbital radius about the Sun in meters and µ = the planet's gravitational constant = Gm where G = the gravitational constant of the Sun and m = the mass of the planet.

To find the mass of the sun, we can use the formula for the centripetal force acting on the planet.

The formula for the centripetal force (F) is:

F = (mass of the planet) x (speed of the planet)^2 / (distance from the sun)

We know the mass of the planet (7 x 10^24 kg), the speed of the planet (43000 m/s), and the period of the planet's orbit (1 Earth year). From the period, we can calculate the distance from the sun using Kepler's third law.

Kepler's third law states that the period of a planet's orbit squared is proportional to the cube of the average distance from the sun:

T^2 = (4π^2 / G x Ms) x R^3

Where T is the period, G is the gravitational constant, Ms is the mass of the sun, and R is the distance from the sun.

Since the period is given as 1 Earth year (which is roughly 365.25 days), we can substitute these values into the equation and solve for R:

(1 Earth year)^2 = (4π^2 / G x Ms) x R^3

(365.25 days)^2 = (4π^2 / G x Ms) x R^3

Once we know the distance from the sun, we can calculate the centripetal force using the given mass and speed of the planet. Rearranging the formula for centripetal force, we have:

F = (mass of the planet) x (speed of the planet)^2 / (distance from the sun)

And since gravitational force provides the centripetal force:

F = (gravitational constant) x (mass of the planet) x (mass of the sun) / (distance from the sun)^2

We can now equate these two expressions for the centripetal force and solve for the mass of the sun.