# 13.1 Newton's law of universal gravitation  (Page 3/6)

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## A collision in orbit

Consider two nearly spherical Soyuz payload vehicles, in orbit about Earth, each with mass 9000 kg and diameter 4.0 m. They are initially at rest relative to each other, 10.0 m from center to center. (As we will see in Kepler’s Laws of Planetary Motion , both orbit Earth at the same speed and interact nearly the same as if they were isolated in deep space.) Determine the gravitational force between them and their initial acceleration. Estimate how long it takes for them to drift together, and how fast they are moving upon impact.

## Strategy

We use Newton’s law of gravitation to determine the force between them and then use Newton’s second law to find the acceleration of each. For the estimate , we assume this acceleration is constant, and we use the constant-acceleration equations from Motion along a Straight Line to find the time and speed of the collision.

## Solution

The magnitude of the force is

$|{\stackrel{\to }{F}}_{12}|={F}_{12}=G\phantom{\rule{0.1em}{0ex}}\frac{{m}_{1}{m}_{2}}{{r}^{2}}=6.67\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-11}\phantom{\rule{0.2em}{0ex}}\text{N}·{\text{m}}^{2}{\text{/kg}}^{2}\frac{\left(9000\phantom{\rule{0.2em}{0ex}}\text{kg}\right)\left(9000\phantom{\rule{0.2em}{0ex}}\text{kg}\right)}{{\left(10\phantom{\rule{0.2em}{0ex}}\text{m}\right)}^{2}}=5.4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\text{N.}$

The initial acceleration of each payload is

$a=\frac{F}{m}=\frac{5.4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}\phantom{\rule{0.2em}{0ex}}\text{N}}{9000\phantom{\rule{0.2em}{0ex}}\text{kg}}=6.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-9}{\text{m/s}}^{2}.$

The vehicles are 4.0 m in diameter, so the vehicles move from 10.0 m to 4.0 m apart, or a distance of 3.0 m each. A similar calculation to that above, for when the vehicles are 4.0 m apart, yields an acceleration of $3.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-8}\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}$ , and the average of these two values is $2.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-8}\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}$ . If we assume a constant acceleration of this value and they start from rest, then the vehicles collide with speed given by

${v}^{2}={v}_{0}^{2}+2a\left(x-{x}_{0}\right),\phantom{\rule{0.2em}{0ex}}\text{where}\phantom{\rule{0.2em}{0ex}}{v}_{0}=0,$

so

$v=\sqrt{2\left(2.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-9}\phantom{\rule{0.2em}{0ex}}\text{N}\right)\left(3.0\phantom{\rule{0.2em}{0ex}}\text{m}\right)}=3.6\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\phantom{\rule{0.2em}{0ex}}\phantom{\rule{0.2em}{0ex}}\text{m/s.}$

We use ${v}^{}={v}_{0}+at$ to find $t=v\text{/}a=1.7\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{s}$ or about 4.6 hours.

## Significance

These calculations—including the initial force—are only estimates, as the vehicles are probably not spherically symmetrical. But you can see that the force is incredibly small. Astronauts must tether themselves when doing work outside even the massive International Space Station (ISS), as in [link] , because the gravitational attraction cannot save them from even the smallest push away from the station.

Check Your Understanding What happens to force and acceleration as the vehicles fall together? What will our estimate of the velocity at a collision higher or lower than the speed actually be? And finally, what would happen if the masses were not identical? Would the force on each be the same or different? How about their accelerations?

The force of gravity on each object increases with the square of the inverse distance as they fall together, and hence so does the acceleration. For example, if the distance is halved, the force and acceleration are quadrupled. Our average is accurate only for a linearly increasing acceleration, whereas the acceleration actually increases at a greater rate. So our calculated speed is too small. From Newton’s third law (action-reaction forces), the force of gravity between any two objects must be the same. But the accelerations will not be if they have different masses.

The effect of gravity between two objects with masses on the order of these space vehicles is indeed small. Yet, the effect of gravity on you from Earth is significant enough that a fall into Earth of only a few feet can be dangerous. We examine the force of gravity near Earth’s surface in the next section.

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