You must be logged in to see worked solutions.

Already have an account? Log in here.

Believe it or not — the world does not revolve around you. Accept this harsh truth, then calculate the beguiling dance of objects in orbit, from binary stars to the symphony of our Solar System.

The distance of a \( m = 9.00 \times 10 ^6 \text{ kg} \) planet from the Sun is \( R = 2.00 \times 10^9 \text{ km}. \) If we stopped the planet in orbit and then let it fall straight towards the Sun, approximately how long would it take for the planet to reach the Sun in seconds?

Consider both the planet and the Sun as point masses.

The mass of the Sun is \( M = 2.00 \times 10^{30} \text{ kg}. \)

Gravitational constant is \( G = 6.67 \times 10^{-11} \text{ Nm}^2\text{/kg}^2. \)

You must be logged in to see worked solutions.

Already have an account? Log in here.

A satellite, moving in an elliptical orbit, is \( 520 \text{ km} \) above Earth's surface at its farthest point and \( 260 \text{ km} \) above at its closest point. Calculate the semimajor axis of the orbit.

The radius of Earth is \( R_e = 6.37 \times 10^6 \text{ m}. \)

You must be logged in to see worked solutions.

Already have an account? Log in here.

A \( 5 \times 10^4 \text{ kg} \) geo-stationary satellite revolves around the earth in a circular orbit of radius \( 36000 \text{ km}. \) Approximately what will be the time period of a \( 5 \times 10^4 \text{ kg} \) satellite in an orbit of radius \( 12000 \text{ km}? \)

A geo-stationary satellite is a satellite whose position in the sky remains the same in the view of a stationary observer on earth.

You must be logged in to see worked solutions.

Already have an account? Log in here.

** _______**.

You must be logged in to see worked solutions.

Already have an account? Log in here.

You must be logged in to see worked solutions.

Already have an account? Log in here.

×

Problem Loading...

Note Loading...

Set Loading...