# 6.5 Newton’s universal law of gravitation  (Page 7/11)

 Page 7 / 11

## 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 $G=6\text{.}\text{674}×{\text{10}}^{\text{–11}}\phantom{\rule{0.25em}{0ex}}\text{N}\cdot {\text{m}}^{2}{\text{/kg}}^{2}$ .

• 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 center 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) $2.937×{\text{10}}^{\text{17}}\phantom{\rule{0.25em}{0ex}}\text{kg}$

b) $4.91×{\text{10}}^{\text{–8}}$

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.

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