## Wednesday, October 16, 2013

### Gravity

Over the weekend, my family and I went to see the movie "Gravity" in IMAX 3D after hearing all the buzz about it.  Admittedly there are some scientific errors in the movie (my son was whispering to me during the movie why Clooney needs to detach himself from Bullock to save her), the movie was a thrilling ride and made you feel what it is like to be adrift in space.  I have always been fascinated by spaceflight ever since I read the books "The right stuff" and "2001: a space odyssey" (I have also seen the movies, of course) as a boy.  The movie also reminds me of a common misconception we have when we see footage of astronauts in space: that there is no gravity in space (hence the term "zero gravity").  We see that as soon as the spacecraft escapes Earth's atmosphere, things like pens start to float around in the capsule.  While is it true that in deep space, far away from other celestial bodies, the effect of their gravitational pull can be minuscule, all manned spaceflights are still relatively close to Earth and will feel its gravitational pull.
Consider astronauts at the International Space Station (ISS).  They experience a gravitational pull similar in strength to what we experience on the ground. Recall Sir Isaac Newton's celebrated equation describing the forces between two bodies:
$$F = \frac{Gm_1m_2}{r^2}$$
where $G\approx 6.67384 \times 10^{-14}m^3g^{-1}s^{-2}$.
Using Newton's second law of motion $F=ma$, the acceleration an object is subjected to due to another object with mass $m$ and a distance $r$ away is $\frac{Gm}{r^2}$.  To calculate the acceleration induced by the gravitational force you feel from the Earth, if we assume the Earth is a sphere with a spherically symmetric density, then $r$ is the distance between you and the center of the Earth, approximately 6.371 Mm. The mass of the Earth $M_e$ is approximately $5.972\times 10^{27}g$, resulting in a constant $\frac{G M_e}{r^2} \approx 9.82 ms^{-2}$.
The ISS is about 0.37 Mm above the ground, increasing $r$ to be about 6.731 Mm.  The resulting constant is then approximately $8.8$, a  10% reduction in weight for everyone on board.  However, the astronauts do not feel this force as they are orbiting the Earth, as they are essentially in continuous free fall.   In other words, without gravity, the spacecraft would move in a straight line, and the speed of the spacecraft is such that the acceleration needed to change the trajectory to an orbit around the Earth is exactly equal to the acceleration induced by the gravitational force.  It is a nice exercise in calculus (which by the way was also invented by Isaac Newton through his introduction of Fluxions and Fluents) to show this.

Since everything is relative, we must also note that we (along with the Earth) are orbiting the Sun as well at a speed of over 100Mm/hr. The distance from the Earth to the Sun is approximately $1.496\times 10^{11}m$ and the mass of the Sun is approximately $1.989\times 10^{33}g$.  Applying Newton's formula above, we get  $\frac{G M_s}{r^2} \approx 0.00593$, which is much smaller than the constant for Earth. Since we and the Earth are in the same orbit (and thus in free fall) around the Sun, we do not notice the effects of the Sun's gravitational pull.

When I told my wife that we are moving around the Sun at high speed, she quipped "So we don't have a sedentary lifestyle then."