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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

| F → 12 | = F 12 = G m 1 m 2 r 2 = 6.67 × 10 −11 N · m 2 /kg 2 ( 9000 kg ) ( 9000 kg ) ( 10 m ) 2 = 5.4 × 10 −5 N.

The initial acceleration of each payload is

a = F m = 5.4 × 10 −5 N 9000 kg = 6.0 × 10 −9 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 × 10 −8 m/s 2 , and the average of these two values is 2.2 × 10 −8 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 + 2 a ( x − x 0 ) , where v 0 = 0 ,

so

v = 2 ( 2.2 × 10 −9 N ) ( 3.0 m ) = 3.6 × 10 −4 m/s.

We use v = v 0 + a t to find t = v / a = 1.7 × 10 4 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.

A photo of an astronaut on a spacewalk is shown.
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.

Got questions? Get instant answers now!

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

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
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Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
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"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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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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