# 13.1 Newton's law of universal gravitation  (Page 4/6)

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## Attraction between galaxies

Find the acceleration of our galaxy, the Milky Way , due to the nearest comparably sized galaxy, the Andromeda galaxy ( [link] ). The approximate mass of each galaxy is 800 billion solar masses (a solar mass is the mass of our Sun), and they are separated by 2.5 million light-years. (Note that the mass of Andromeda is not so well known but is believed to be slightly larger than our galaxy.) Each galaxy has a diameter of roughly 100,000 light-years $\left(1\phantom{\rule{0.2em}{0ex}}\text{light-year}=9.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{15}\phantom{\rule{0.2em}{0ex}}\phantom{\rule{0.2em}{0ex}}\text{m}\right)$ .

## Strategy

As in the preceding example, 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 the Milky Way. We can consider the galaxies to be point masses, since their sizes are about 25 times smaller than their separation. The mass of the Sun (see Appendix D ) is $2.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{30}\phantom{\rule{0.2em}{0ex}}\text{kg}$ and a light-year is the distance light travels in one year, $9.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{15}\phantom{\rule{0.2em}{0ex}}\text{m}$ .

## Solution

The magnitude of the force is

${F}_{12}=G\phantom{\rule{0.1em}{0ex}}\frac{{m}_{1}{m}_{2}}{{r}^{2}}=\left(6.67\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-11}\phantom{\rule{0.2em}{0ex}}\text{N}·{\text{m}}^{2}\text{/}{\text{kg}}^{2}\right)\frac{{\left[\left(800\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{9}\right)\left(2.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{30}\phantom{\rule{0.2em}{0ex}}\text{kg}\right)\right]}^{2}}{{\left[\left(2.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{6}\right)\left(9.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{15}\phantom{\rule{0.2em}{0ex}}\text{m}\right)\right]}^{2}}=3.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{29}\phantom{\rule{0.2em}{0ex}}\text{N.}$

The acceleration of the Milky Way is

$a=\frac{F}{m}=\frac{3.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{29}\phantom{\rule{0.2em}{0ex}}\text{N}}{\left(800\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{9}\right)\left(2.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{30}\phantom{\rule{0.2em}{0ex}}\text{kg}\right)}=1.9\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-13}{\text{m/s}}^{2}.$

## Significance

Does this value of acceleration seem astoundingly small? If they start from rest, then they would accelerate directly toward each other, “colliding” at their center of mass. Let’s estimate the time for this to happen. The initial acceleration is $~{10}^{-13}\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}$ , so using $v=at$ , we see that it would take $~{10}^{13}\phantom{\rule{0.2em}{0ex}}\text{s}$ for each galaxy to reach a speed of 1.0 m/s, and they would be only $~0.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{13}\phantom{\rule{0.2em}{0ex}}\text{m}$ closer. That is nine orders of magnitude smaller than the initial distance between them. In reality, such motions are rarely simple. These two galaxies, along with about 50 other smaller galaxies, are all gravitationally bound into our local cluster. Our local cluster is gravitationally bound to other clusters in what is called a supercluster. All of this is part of the great cosmic dance that results from gravitation, as shown in [link] .

## Summary

• All masses attract one another with a gravitational force proportional to their masses and inversely proportional to the square of the distance between them.
• Spherically symmetrical masses can be treated as if all their mass were located at the center.
• Nonsymmetrical objects can be treated as if their mass were concentrated at their center of mass, provided their distance from other masses is large compared to their size.

## Conceptual questions

Action at a distance, such as is the case for gravity, was once thought to be illogical and therefore untrue. What is the ultimate determinant of the truth in science, and why was this action at a distance ultimately accepted?

The ultimate truth is experimental verification. Field theory was developed to help explain how force is exerted without objects being in contact for both gravity and electromagnetic forces that act at the speed of light. It has only been since the twentieth century that we have been able to measure that the force is not conveyed immediately.

In the law of universal gravitation, Newton assumed that the force was proportional to the product of the two masses ( $~{m}_{1}{m}_{2}$ ). While all scientific conjectures must be experimentally verified, can you provide arguments as to why this must be? (You may wish to consider simple examples in which any other form would lead to contradictory results.)

## Problems

Evaluate the magnitude of gravitational force between two 5-kg spherical steel balls separated by a center-to-center distance of 15 cm.

$7.4\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-8}\phantom{\rule{0.2em}{0ex}}\text{N}$

Estimate the gravitational force between two sumo wrestlers, with masses 220 kg and 240 kg, when they are embraced and their centers are 1.2 m apart.

Astrology makes much of the position of the planets at the moment of one’s birth. The only known force a planet exerts on Earth is gravitational. (a) Calculate the gravitational force exerted on a 4.20-kg baby by a 100-kg father 0.200 m away at birth (he is assisting, so he is close to the child). (b) Calculate the force on the baby due to Jupiter if it is at its closest distance to Earth, some $6.29\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{11}\phantom{\rule{0.2em}{0ex}}\text{m}$ away. How does the force of Jupiter on the baby compare to the force of the father on the baby? Other objects in the room and the hospital building also exert similar gravitational forces. (Of course, there could be an unknown force acting, but scientists first need to be convinced that there is even an effect, much less that an unknown force causes it.)

a. $7.01\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7}\phantom{\rule{0.2em}{0ex}}\text{N}$ ; b. The mass of Jupiter is
$\begin{array}{}\\ \\ {m}_{\text{J}}=1.90\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{27}\phantom{\rule{0.2em}{0ex}}\text{kg}\hfill \\ {F}_{\text{J}}=1.35\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}\text{N}\hfill \\ \frac{{F}_{\text{f}}}{{F}_{\text{J}}}=0.521\hfill \end{array}$

A mountain 10.0 km from a person exerts a gravitational force on him equal to 2.00% of his weight. (a) Calculate the mass of the mountain. (b) Compare the mountain’s mass with that of Earth. (c) What is unreasonable about these results? (d) Which premises are unreasonable or inconsistent? (Note that accurate gravitational measurements can easily detect the effect of nearby mountains and variations in local geology.)

The International Space Station has a mass of approximately 370,000 kg. (a) What is the force on a 150-kg suited astronaut if she is 20 m from the center of mass of the station? (b) How accurate do you think your answer would be?

a. $9.25\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-6}\phantom{\rule{0.2em}{0ex}}\text{N}$ ; b. Not very, as the ISS is not even symmetrical, much less spherically symmetrical.

Asteroid Toutatis passed near Earth in 2006 at four times the distance to our Moon. This was the closest approach we will have until 2060. If it has mass of $5.0\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{13}\phantom{\rule{0.2em}{0ex}}\text{kg}$ , what force did it exert on Earth at its closest approach?

(a) What was the acceleration of Earth caused by asteroid Toutatis (see previous problem) at its closest approach? (b) What was the acceleration of Toutatis at this point?

a. $1.41\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-15}{\text{m/s}}^{2}$ ; b. $1.69\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-4}\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}$

A 10kg ball travelling at 4meter per second collides elastically in a head-on collision with a 2kg ball.What are (a)the velocities and (b)the total momentum of the balls after collision?
The displacement of the air molecules in sound wave is modeled with the wave function s(x,t)=5.00nmcos(91.54m−1x−3.14×104s−1t)s(x,t)=5.00nmcos(91.54m−1x−3.14×104s−1t) . (a) What is the wave speed of the sound wave? (b) What is the maximum speed of the air molecules as they oscillate in simple harmon
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Luminous
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f=ma
F=m×a Where F=force M=mass of a body of an object a=acceleration due to gravity
Abubakar
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distance travelled per unit of time is speed.
distance travelled in a particular direction it is.
Andrew
Speed is define as the distance move per unit time. Mathematically is given as Speed = distance/time Speed = s/t
Abubakar
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v=ms^-1 velocity=distance time
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velocity= displacement time
Gold
Velocity = speed/time
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If the block is displaced to a position y , the net force becomes Fnet=k(y−y0)−mg=0Fnet=k(y−y0)−mg=0 . But we found that at the equilibrium position, mg=kΔy=ky0−ky1mg=kΔy=ky0−ky1 . Substituting for the weight in the equation yields. Show me an equation of graph.
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Is equal to the square of the velocity divided by the radius of circular path of the object
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A Radial Acceleration is defined as the upward movement of an object.
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I believe because speed is a function of air density, and colder air is more dense
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At night air is denser because of humidity.
Clifton
Night air is cooler. Sound requires medium to travel so the denser the medium the fastest the sound travels. Humid air is denser then warmer air as in day.
Clifton
The humidity statement is misleading , colder air is more dense period.
Jerry
because there is no any other sound to reverberate with it so it clearly travel to lot of distance and also humidity and also due to denser air at night
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because it is quiet at night. this takes us to the topic wave, it depends on the wave at that moment, which Echo's....sound travelled.
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element radioactivit diffusion atomic radius alpha beta gamma these elements have more than 92 atomic mass start from uranium
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alpha discoved by ratherforth
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alpha discoved by Rutherford in 1899 have + charge and similarly to helium nucleus . When give 2 electron it change to helium nuclear( alpha + 2e = he
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