Between the orbits of Mars and Jupiter, several thousand small objects called asteroids move in nearly circular orbits around the Sun. Consider an asteroid that is spherically shaped with radius r and density 2400 kg/m3 .

Part A) You find yourself on the surface of this asteroid and throw a baseball at a speed of 30 m/s . If the baseball is to travel around the asteroid in a circular orbit, what is the largest radius asteroid on which you are capable of accomplishing this feat?

Part B) After you throw the baseball, you turn around and face the opposite direction and catch the baseball. How much time T elapses between your throw and your catch?

Respuesta :

Newton's second law and universal gravitation law we can the answers are:

     a) r = 7.3 10⁴ m

     b) T = 1.53 10⁴ s

 

Newton's second law states that force is proportional to slats and the acceleration of the body,

            F = m a

The universal gravitational law says that the force between two bodies is proportional to their plus and inversely proportional to the distance between them

            F = [tex]G \frac{Mm}{r^2}[/tex]

Where the bold letters indicate vectors and F is the force, G the universal gravitation constant (G = 6.67 10⁻¹¹ N m² /kg²), M the most of the asteroid, m the most of the ball and r the separation between the center of the asteroid and the ball

             

a) The acceleration is centripetal since the ball rotates around the asteroid in a circular orbit

           a = v² / r

We substitute

            [tex]G \frac{Mm}{r^2} = m \frac{v^2}{r}[/tex]G Mm / r² = m v² / r

            [tex]G \frac{M}{r}[/tex] = v²

 

Density is defined by the relationship between the mass and the volume of a body

         ρ = [tex]\frac{M}{V}[/tex]M / V

         M = ρ V

In this case it indicates that the asteroid is circular, so the volume (V) of a sphere is

         V = [tex]\frac{4}{3} \pi r^3[/tex]

          M = [tex]\rho \ \frac{4}{3} \pi r^3[/tex]

In this case we have assumed that the distance to the ball and the radius of the asteroid are equal

we substitute

           [tex]\frac{4}{3} \pi \ G\ \rho r^2 = v^2[/tex]

           r² = [tex]\frac{3}{4} \ \frac{v^2} {\pi \ G \ \rho }[/tex]i

Let's calculate

           r² = ¾  [tex]\frac{30^2}{ \pi \ 6.67 \ 10^{-11} \ 2400}[/tex]

           r = [tex]\sqrt{ 0.536879 \ 10^{10}}[/tex]

           r = 0.7327 10⁵ m

           r = 7.3 10⁴ m

b) As the motion of the ball is circulating, the modulus of its velocity is constant, for which we can use the kinematic relations of the uniform motion

        v = d / T

        T = d / v

Where d is the distance and T the period of the orbit

The distance of a circle is

        d = 2π r

        d = 2π 7.3 10⁴

        d = 4.60 10⁵ m

we calculate

       T = 46.0 10⁴ / 30

        T = 1.53 10⁴ s

In conclusion using Newton's second law and the universal law of gravitation we can the answers are:

     a) r = 7.3 10⁴ m

     b) T = 1.53 10⁴ s

Learn more about the universal gravitation law here:

https://brainly.com/question/858421

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