# 6.3 Newton’s universal law of gravitation  (Page 6/10)

 Page 6 / 10

## Section summary

• Newton’s universal law of gravitation: Every particle in the universe attracts every other particle with a force along a line joining them. The force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. In equation form, this is
$F=G\frac{\text{mM}}{{r}^{2}}\text{,}$

where F is the magnitude of the gravitational force. $G$ is the gravitational constant, given by $G=6\text{.}\text{673}×{\text{10}}^{\text{–11}}\phantom{\rule{0.25em}{0ex}}\text{N}\cdot {\text{m}}^{2}{\text{/kg}}^{2}$ .

• Newton’s law of gravitation applies universally.

## 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 physics, and why was this action ultimately accepted?

Two friends are having a conversation. Anna says a satellite in orbit is in freefall because the satellite keeps falling toward Earth. Tom says a satellite in orbit is not in freefall because the acceleration due to gravity is not $9.80 m{\text{/s}}^{2}$ . Who do you agree with and why?

Draw a free body diagram for a satellite in an elliptical orbit showing why its speed increases as it approaches its parent body and decreases as it moves away.

Newton’s laws of motion and gravity were among the first to convincingly demonstrate the underlying simplicity and unity in nature. Many other examples have since been discovered, and we now expect to find such underlying order in complex situations. Is there proof that such order will always be found in new explorations?

## Problem exercises

(a) Calculate Earth’s mass given the acceleration due to gravity at the North Pole is $9.830 m{\text{/s}}^{2}$ and the radius of the Earth is 6371 km from pole to pole.

(b) Compare this with the accepted value of $5\text{.}\text{979}×{\text{10}}^{\text{24}}\phantom{\rule{0.25em}{0ex}}\text{kg}$ .

a) $5.979×{\text{10}}^{\text{24}}\phantom{\rule{0.25em}{0ex}}\text{kg}$

b) This is identical to the best value to three significant figures.

(a) Calculate the magnitude of the acceleration due to gravity on the surface of Earth due to the Moon.

(b) Calculate the magnitude of the acceleration due to gravity at Earth due to the Sun.

(c) Take the ratio of the Moon’s acceleration to the Sun’s and comment on why the tides are predominantly due to the Moon in spite of this number.

(a) What is the acceleration due to gravity on the surface of the Moon?

(b) On the surface of Mars? The mass of Mars is $6.418×{\text{10}}^{\text{23}}\phantom{\rule{0.25em}{0ex}}\text{kg}$ and its radius is $3\text{.}\text{38}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{m}$ .

a) $1.62\phantom{\rule{0.5em}{0ex}}\text{m}/{\text{s}}^{2}$

b) $3.75\phantom{\rule{0.5em}{0ex}}\text{m}/{\text{s}}^{2}$

(a) Calculate the acceleration due to gravity on the surface of the Sun.

(b) By what factor would your weight increase if you could stand on the Sun? (Never mind that you cannot.)

The Moon and Earth rotate about their common center of mass, which is located about 4700 km from the center of Earth. (This is 1690 km below the surface.)

(a) Calculate the magnitude of the acceleration due to the Moon’s gravity at that point.

(b) Calculate the magnitude of the centripetal acceleration of the center of Earth as it rotates about that point once each lunar month (about 27.3 d) and compare it with the acceleration found in part (a). Comment on whether or not they are equal and why they should or should not be.

a) $3.42×{\text{10}}^{\text{–5}}\phantom{\rule{0.25em}{0ex}}\text{m}/{\text{s}}^{2}$

b) $3.34×{\text{10}}^{\text{–5}}\phantom{\rule{0.25em}{0ex}}\text{m}/{\text{s}}^{2}$

The values are nearly identical. One would expect the gravitational force to be the same as the centripetal force at the core of the system.

Solve part (b) of [link] using ${a}_{c}={v}^{2}/r$ .

Astrology, that unlikely and vague pseudoscience, 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 magnitude of 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 magnitude of the force on the baby due to Jupiter if it is at its closest distance to Earth, some $6\text{.}\text{29}×{\text{10}}^{\text{11}}\phantom{\rule{0.25em}{0ex}}\text{m}$ 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) $7.01×{\text{10}}^{\text{–7}}\phantom{\rule{0.25em}{0ex}}\text{N}$

b) $1.35×{\text{10}}^{\text{–6}}\phantom{\rule{0.25em}{0ex}}\text{N}$ , $0.521$

The existence of the dwarf planet Pluto was proposed based on irregularities in Neptune’s orbit. Pluto was subsequently discovered near its predicted position. But it now appears that the discovery was fortuitous, because Pluto is small and the irregularities in Neptune’s orbit were not well known. To illustrate that Pluto has a minor effect on the orbit of Neptune compared with the closest planet to Neptune:

(a) Calculate the acceleration due to gravity at Neptune due to Pluto when they are $4\text{.}\text{50}×{\text{10}}^{\text{12}}\phantom{\rule{0.25em}{0ex}}\text{m}$ apart, as they are at present. The mass of Pluto is $1\text{.}4×{\text{10}}^{\text{22}}\phantom{\rule{0.25em}{0ex}}\text{kg}$ .

(b) Calculate the acceleration due to gravity at Neptune due to Uranus, presently about $2\text{.}\text{50}×{\text{10}}^{\text{12}}\phantom{\rule{0.25em}{0ex}}\text{m}$ apart, and compare it with that due to Pluto. The mass of Uranus is $8\text{.}\text{62}×{\text{10}}^{\text{25}}\phantom{\rule{0.25em}{0ex}}\text{kg}$ .

(a) The Sun orbits the Milky Way galaxy once each $2\text{.}{\text{60 x 10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{y}$ , with a roughly circular orbit averaging $3\text{.}{\text{00 x 10}}^{4}$ light years in radius. (A light year is the distance traveled by light in 1 y.) Calculate the centripetal acceleration of the Sun in its galactic orbit. Does your result support the contention that a nearly inertial frame of reference can be located at the Sun?

(b) Calculate the average speed of the Sun in its galactic orbit. Does the answer surprise you?

a) $1.66×{\text{10}}^{\text{–10}}\phantom{\rule{0.25em}{0ex}}\text{m}/{\text{s}}^{2}$

b) $2.17×{\text{10}}^{\text{5}}\phantom{\rule{0.25em}{0ex}}\text{m/s}$

Unreasonable Result

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.)

a) $2.94×{\text{10}}^{\text{17}}\phantom{\rule{0.25em}{0ex}}\text{kg}$

b) $4.92×{\text{10}}^{\text{–8}}$

of the Earth’s mass.

c) The mass of the mountain and its fraction of the Earth’s mass are too great.

d) The gravitational force assumed to be exerted by the mountain is too great.

find the 15th term of the geometric sequince whose first is 18 and last term of 387
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Got questions? Join the online conversation and get instant answers!