Question

An object revolves around the Earth in a circular orbit whose radius is five times that of the radius of the Earth. The gravitational field strength at the surface of the Earth is 9.8 N kg. Determine the period of revolution of the object. ​

Answer 1

To determine the period of revolution of the object, we can use Kepler's third law of planetary motion. The period is given by T = sqrt((4π²/(9.8 N/kg)) * 125r³), where r is the radius of the Earth and the radius of the orbit is five times that of the Earth.

Step-by-step explanation:

To determine the period of revolution of the object, we can use Kepler's third law of planetary motion, which states that the square of the period of revolution is proportional to the cube of the semi-major axis of the orbit. In this case, the semi-major axis is the radius of the orbit, which is five times the radius of the Earth.

Let's denote the period of revolution as T and the radius of the Earth as r. Using the given information, we have:

• The radius of the orbit: 5r
• The gravitational field strength at the surface of the Earth: 9.8 N/kg

Plugging these values into the equation for Kepler's third law, we get:

T^2 = (4π²/(9.8 N/kg)) * (5r)³

Simplifying the equation, we find:

T^2 = (4π²/(9.8 N/kg)) * 125r³

And taking the square root of both sides, we get:

T = sqrt((4π²/(9.8 N/kg)) * 125r³)

Therefore, the period of revolution of the object is given by the equation above.