Question

A satellite orbiting in a circular orbit with a radius of 66000 m and a speed of 810 m/s, fires a rocket that provides an acceleration of 25 m/s/s. At the instant the rocket fires, what are the magnitude and direction of the total acceleration

Answer 1

Answer:

The magnitude and direction of the total acceleration is 34.941m/s² upward

Explanation:

The centripetal acceleration due to the circular motion, is calculated as follows;

$a =\frac{v^2}{r}$

where;

a is the centripetal acceleration

v is the speed of the satellite =  810 m/s

r is the radius of the circular orbit = 66000 m

$a = \frac{810^2}{66000} =9.941\frac{m}{s^2}$

Upward acceleration = 25m/s²

Total acceleration = (9.941 + 25)m/s² = 34.941m/s² upward

Therefore, the magnitude and direction of the total acceleration is 34.941m/s² upward