Assume you are agile enough to run across a horizontal surface at 8.10 m/s on an airless...


Assume you are agile enough to run across a horizontal surface at 8.10 m/s on an airless spherical asteroid of uniform density 1.9x103 kg/m3, independently of the value of the gravitational field. To launch yourself into orbit by running, what would be (a) the radius?

Orbital Motion:

Orbital motion of a satellite around planet is possible just because of the gravitational force between the planet and satellite. If this gravitational force is sufficient enough to provide centripetal force for the satellite in definite orbit at some definite distance from the center of the planet.

The gravitational force exerted by the planet of mass M on the satellite of mass m at a distance r from the center of the planet is {eq}F= \dfrac{ G M m } { r^2 } {/eq}.

Here G is the gravitational constant. In case if we know only the density {eq}\rho {/eq} of the planet and its radius R, the mass of the planet can be expressed as {eq}M = \dfrac { 4 } { 3 } \pi R^3 \rho {/eq}.

The planet can have a satellite of mass m at some orbit of radius r with some orbital velocity v such that the centripetal force {eq}F_c = \dfrac { m v^2 } { r } {/eq} equal to gravitational force.

Answer and Explanation:

Given data

  • Maximum speed with which the person can run on the surface of the asteroid {eq}v = 8.10 \ m/s {/eq}
  • Density of the asteroid {eq}\rho = 1.9 \times 10^3 \ kg/ m^3 {/eq}
  • Gravitational constant {eq}G = 6.674 \times 10^{-11} \ m^3 / kg . s^2 {/eq}

Let {eq}M, \ \ m {/eq} be the mass of the asteroid and the person respectively.

Let R be the radius of the asteroid.

The person can launch into an orbit just above the surface of the asteroid if the gravitational force can provide the centripetal force.

Let us assume the the gravitational force of the asteroid can provide centripetal force for an orbit of radius R just equal to that of the radius itself.

In that case we get {eq}\dfrac { m v^2 } { R } = \dfrac { G M m } { R^2 } \\ \implies v^2 = \dfrac { G M } { R } {/eq}

The mass of the asteroid in terms radius and density {eq}M = \dfrac { 4 } { 3 } \pi R^3 \times \rho {/eq}

Using this in the above equation we get {eq}v^2 = \dfrac { G \times 4 \pi R^2 \times \rho } { 3 } {/eq}

Therefore radius of the asteroid {eq}R = \sqrt { \dfrac { 3 v^2 } { 4 G \pi \rho } }\\ R = \sqrt { \dfrac { 3 \times 8.10^2 } { 4 \times 6.674 \times 10^{-11} \times \pi \times 1.9 \times 10^3 } } \\ R = 1.11 \times 10^4 \ m {/eq}

Learn more about this topic:

Stable Orbital Motion of a Satellite: Physics Lab

from Physics: High School

Chapter 8 / Lesson 16

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