Question

The gravity tractor is a proposed spacecraft that will fly close to an asteroid whose trajectory threatens to impact the Earth. Due to the gravitational attraction between the two objects and a prolonged period of time over which it acts (several years), the asteroid's trajectory is changed slightly, thus hopefully diverting it from impacting the Earth. If the gravity tractor's weight on earth is 22000 lb and it flies with its center of gravity 150 ft from the surface of the asteroid, and the asteroid is homogeneous pure iron with 1030 ft diameter spherical shape, determine the force of mutual attraction. Idealize the gravity tractor to be a particle.

Answer 1

Answer:

The force of mutual attraction between the asteroid and the gravity tractor is 0.465 lbf

Explanation:

1. In earth, a body that has a weight of 22000 lbf, would have a mass of 22000 lbm.
2. The asteroid, if is an sphere, would have a volume of $V=\frac{4}{3} \pi r^{3}=\frac{4}{3} \pi(\frac{1030ft}{2})^{3} =572150519.3ft^3$
3. The iron has a density of $491.616\frac{lbm}{ft^{3} }$, then the mass of the asteroid would be $mass=density*volume=491.616\frac{lbm}{ft^{3} } *572150519.3ft^{3}=2.81278*10^{11}lbm$
4. If the distance between the surface of the asteroid and the gravity tractor is 150 ft, and the diameter of the asteroid is 1030 ft, then the distance between the tractor and the center of the asteroid would be = $\frac{1030ft}{2} +150ft=665ft$
5. The force of attraction between two bodies is described by the Newton's law of universal gravitation: $F=G\frac{m_{1}* m_{2} }{r^{2} }$, where $F$ is the force of attraction, $m_{1}$ and $m_{2}$ are the masses of each body, $r$ is the distance between both bodies, and $G= 1.068846*10^{-9}\frac{ft^{3} }{lb*s^{2} }$ is called the gravitational constant.
6. Then if it is supposed that the mass is concenter in the center of each body. The force of attraction would be: $F=1.068846*10^{-9}\frac{ft^{3} }{lb*s^{2} } \frac{22000lbm*2.813*10^{11}lbm }{(665ft)^{2} } =14.96\frac{lbm*ft}{s^{2}} \frac{1lbf}{32.17lbm*\frac{ft}{s^{2} } } =0.4649lbf$