# 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. This photo shows Ed White tethered to the Space Shuttle during a spacewalk. (credit: NASA)

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.

#### Questions & Answers

Suppose the master cylinder in a hydraulic system is at a greater height than the cylinder it is controlling. Explain how this will affect the force produced at the cylinder that is being controlled.
Louise Reply
Why is popo less than atmospheric? Why is popo greater than pipi?
Louise
The old rubber boot shown below has two leaks. To what maximum height can the water squirt from Leak 1? How does the velocity of water emerging from Leak 2 differ from that of Leak 1? Explain your responses in terms of energy.
Louise
David rolled down the window on his car while driving on the freeway. An empty plastic bag on the floor promptly flew out the window. Explain why.
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the pressure differential exerted a force on the bag greater than the gravitational force holding it on the floor.
gtitboi
what is angular velocity
Sthandazile Reply
The rate of change in angular displacement is defined as angular velocity.
Manorama
a length of copper wire was measured to be 50m with an uncertainty of 1cm, the thickness of the wire was measured to be 1mm with an uncertainty of 0.01mm, using a micrometer screw gauge, calculate the of copper wire used
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What is the answer please
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If centripetal force is directed towards the center,why do you feel that you're thrown away from the center as a car goes around a curve? Explain
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if there is a centripetal force it means that there's also a centripetal acceleration, getting back to your question, just imagine what happens if you pull out of a car when it's quickly moving or when you try to stop when you are running fast, anyway, we notice that there's always a certain force..
Lindomar
... that tends to fight for its previous direction when you try to attribute to it an opposite one ou try to stop it.The same thing also happens whe a car goes around a curve, the car it self is designed to a"straight line"(look at the position of its tyres, mainly the back side ones), so...
Lindomar
... whenever it goes around a curve, it tends to throw away its the occupiers, it's given to the fact that it must interrupt its initial direction and take a new one.
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Which kind of wave does wind form
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calculate the distance you will travel if you mantain an average speed of 10N m/s for 40 second
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400m/s
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2.5
omwoyo
2.54cm=1inche
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if m2 is twice of m1. find the ration of kinetic energy in COM system to lab system of elastic collision
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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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