Question

A satellite moves on a circular earth orbit that has a radius of 6,758,998 m. A model airplane is flying on a 10 m guideline in a horizontal circle. The guideline is parallel to the ground. Find the speed of the plane such that the plane and the satellite have the same centripetal acceleration.

Answer 1

Answer:

v = 0.1068 m / s

Explanation:

To find the speed of the satellite we use Newton's second law where the force is the universal law of gravitation

            F = ma

            F = $G \frac{m M}{r^2}$

acceleration is centripetal

            a = v² / r

we substitute

            $G \frac{m M}{r^2} = m \frac{v^2}{r}$

             v² = $G \frac{M}{r}$

The radius of the orbit is given we will assume that this radius is half from the center of the earth

we substitute

            v² = 6.67 10⁻¹¹ 5.98 10²⁴/6758998

            v = $\sqrt{59.013 \ 10^6}$

            v = 7.68 10³ m / s

The centripetal acceleration is

            a = v² / r

            a = 7.68 10³/6758998

            a = 1.14 10⁻³ m / s²

For the airplane we use the definition of centripetal acceleration

            a = v² / r

            v = $\sqrt{a \ r }$

           

let's calculate

            v = $\sqrt{1.14 \ 10^{-3} \ 10}$

            v = 0.1068 m / s