Question

To teach you how to find the parameters characterizing an object in a circular orbit around a much heavier body like the earth.

Answer 1

Answer:

a)  T² = ($\frac{4\pi ^2}{GM}$)  r³

b) veloicity the dependency is the inverse of the root of the distance

kinetic energy  depends on the inverse of the distance

potential energy dependency is the inverse of distance

angular momentum depends directly on the root of the distance

Explanation:

1) for this exercise we will use Newton's second law

            F = ma

in this case the acceleration is centripetal

            a = v² / r

the linear and angular variable are related

           v = w r

we substitute

           a = w² r

force is the universal force of attraction

           F = $G \frac{m M}{r^2}$

we substitute

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

         w² = $\frac{GM}{r^3}$

angular velocity is related to frequency and period

         w = 2π f = 2π / T

we substitute

            $( \frac{2\pi }{T} ) = \frac{GM}{r^3}$

the final equation is

             T² = ()  r³

b) the speed of the orbit can be found

           v = w r

            v = $\sqrt{\frac{GM}{r^3} } \ r$

            v = $\sqrt{\frac{GM}{r} }$

in this case the dependency is the inverse of the root of the distance

Kinetic energy

           K = ½ M v²

           K = ½ M GM / r

           K = ½ GM² 1 / r

the kinetic energy depends on the inverse of the distance

Potential energy

          U =

          U = -G mM / r

dependency is the inverse of distance

Angular momentum

          L = r x p

for a circular orbit

           L = r p = r Mv

           L =

         L =

The angular momentum depends directly on the root of the distance