Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity

7.2 Newton's Law of Universal Gravitation and Einstein's Theory of General Relativity

Section Learning Objectives

By the end of this section, you will be able to do the following:

Explain Newton’s law of universal gravitation and compare it to Einstein’s theory of general relativity
Perform calculations using Newton’s law of universal gravitation

Teacher Support

The learning objectives in this section will help your students master the following standards:

(4) Science concepts. The student knows and applies the laws governing motion in a variety of situations. The student is expected to:
- (D) calculate the effect of forces on objects, including the law of inertia, the relationship between force and acceleration, and the nature of force pairs between objects;

(5) Science concepts. The student knows the nature of forces in the physical world. The student is expected to:
- (A) research and describe the historical development of the concepts of gravitational, electromagnetic, weak nuclear, and strong nuclear forces;
- (B) describe and calculate how the magnitude of the gravitational force between two objects depends on their masses and the distance between their centers.

Section Key Terms

Einstein’s theory of general relativity

gravitational constant

Newton’s universal law of gravitation

Teacher Support

In this section, students will apply Newton’s law of universal gravitation to objects close at hand and far off in the depths of the solar system.

[BL][OL] Compare the contributions of Kepler, Newton, and Einstein. Place them historically with dates.

[AL] Ask if anyone knows the difference between special relativity and general relativity. Special relativity is a theory of spacetime and applies to observers moving at constant velocity. General relativity is a theory of gravity and applies to observers that are accelerating. General relativity is broader and includes special relativity, which was published first.

Concepts Related to Newton’s Law of Universal Gravitation

Sir Isaac Newton was the first scientist to precisely define the gravitational force, and to show that it could explain both falling bodies and astronomical motions. See Figure 7.7. But Newton was not the first to suspect that the same force caused both our weight and the motion of planets. His forerunner, Galileo Galilei, had contended that falling bodies and planetary motions had the same cause. Some of Newton’s contemporaries, such as Robert Hooke, Christopher Wren, and Edmund Halley, had also made some progress toward understanding gravitation. But Newton was the first to propose an exact mathematical form and to use that form to show that the motion of heavenly bodies should be conic sections—circles, ellipses, parabolas, and hyperbolas. This theoretical prediction was a major triumph. It had been known for some time that moons, planets, and comets follow such paths, but no one had been able to propose an explanation of the mechanism that caused them to follow these paths and not others.

A painting shows a man sitting under a tree, from which an apple is falling.

Figure 7.7 The popular legend that Newton suddenly discovered the law of universal gravitation when an apple fell from a tree and hit him on the head has an element of truth in it. A more probable account is that he was walking through an orchard and wondered why all the apples fell in the same direction with the same acceleration. Great importance is attached to it because Newton’s universal law of gravitation and his laws of motion answered very old questions about nature and gave tremendous support to the notion of underlying simplicity and unity in nature. Scientists still expect underlying simplicity to emerge from their ongoing inquiries into nature.

Teacher Support

[BL][OL]Ask students if it really is obvious why all things fall straight down. Ask them to back up their reasons. Ask if the name Halley rings a bell.

[OL][AL]Ask if anyone thinks it is strange or even mysterious that a force can act at a distance across empty space. Ask the students to compare and contrast gravitational force with magnetic and electrostatic forces. Note how much force at a distance is like magic or having superpowers.

The gravitational force is relatively simple. It is always attractive, and it depends only on the masses involved and the distance between them. Expressed in modern language, Newton’s universal law of gravitation states that every object in the universe attracts every other object with a force that is directed 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. This attraction is illustrated by Figure 7.8.

A figure shows a small circle and a large circle. A black line goes from one circle to the other. A red line with an arrow extends outward from each circle. The red lines have arrows on one end, pointing away from the circles, towards the midpoint of the space in between the circles.

Figure 7.8 Gravitational attraction is along a line joining the centers of mass (CM) of the two bodies. The magnitude of the force on each body is the same, consistent with Newton’s third law (action-reaction).

For two bodies having masses m and M with a distance r between their centers of mass, the equation for Newton’s universal law of gravitation is

F = G \frac{m M}{r^{2}}

where F is the magnitude of the gravitational force and G is a proportionality factor called the gravitational constant. G is a universal constant, meaning that it is thought to be the same everywhere in the universe. It has been measured experimentally to be $G = 6.673 \times 10^{- 11} N \cdot m^{2} {/kg}^{2}$ .

If a person has a mass of 60.0 kg, what would be the force of gravitational attraction on him at Earth’s surface? G is given above, Earth’s mass M is 5.97 × 10²⁴ kg, and the radius r of Earth is 6.38 × 10⁶ m. Putting these values into Newton’s universal law of gravitation gives

F = G \frac{m M}{r^{2}} = (6.673 \times 10^{- 11} \frac{N \cdot m^{2}}{{kg}^{2}}) (\frac{(60.0 kg) (5.97 \times 10^{24} kg)}{{(6.38 \times 10^{6} m)}^{2}}) = 584 N

We can check this result with the relationship: $F = m g = (60 kg) (9.8 {m/s}^{2}) = 588 N$

You may remember that g, the acceleration due to gravity, is another important constant related to gravity. By substituting g for a in the equation for Newton’s second law of motion we get $F = m g$ . Combining this with the equation for universal gravitation gives

m g = G \frac{m M}{r^{2}}

Cancelling the mass m on both sides of the equation and filling in the values for the gravitational constant and mass and radius of the Earth, gives the value of g, which may look familiar.

g = G \frac{M}{r^{2}} = (6.67 \times 10^{- 11} \frac{N \cdot m^{2}}{{kg}^{2}}) (\frac{5.98 \times 10^{24} kg}{{(6.38 \times 10^{6} m)}^{2}}) = 9.80 {m/s}^{2}

This is a good point to recall the difference between mass and weight. Mass is the amount of matter in an object; weight is the force of attraction between the mass within two objects. Weight can change because g is different on every moon and planet. An object’s mass m does not change but its weight mg can.

Teacher Support

[BL][OL]Be sure no one is confusing G with g.

[AL]Ask if anyone can explain why G is a universal constant that applies anywhere in the universe. Have them discuss the idea that the laws of physics are the same everywhere and that, at one time, people were not so sure about this. Emphasize that g is not a universal constant.

Virtual Physics

Gravity and Orbits

Move the sun, Earth, moon and space station in this simulation to see how it affects their gravitational forces and orbital paths. Visualize the sizes and distances between different heavenly bodies. Turn off gravity to see what would happen without it!

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Grasp Check

Would you have the same mass on the moon as you do on Earth? Would you have the same weight?

You would weigh more on the moon than on Earth because gravity on the moon is stronger than gravity on Earth.
You would weigh less on the moon than on Earth because gravity on the moon is weaker than gravity on Earth.
You would weigh less on the moon than on Earth because gravity on the moon is stronger than gravity on Earth.
You would weigh more on the moon than on Earth because gravity on the moon is weaker than gravity on Earth.

Teacher Support

This may be a rather long-winded explanation of the mass-weight distinction, but it should drive home the point.

Two equations involving the gravitational constant, G, are often useful. The first is Newton’s equation, $F = G \frac{m M}{r^{2}}$ . Several of the values in this equation are either constants or easily obtainable. F is often the weight of an object on the surface of a large object with mass M, which is usually known. The mass of the smaller object, m, is often known, and G is a universal constant with the same value anywhere in the universe. This equation can be used to solve problems involving an object on or orbiting Earth or other massive celestial object. Sometimes it is helpful to equate the right-hand side of the equation to mg and cancel the m on both sides.

The equation $\frac{r^{3}}{T^{2}} = \frac{G M}{4 π^{2}}$ is also useful for problems involving objects in orbit. Note that there is no need to know the mass of the object. Often, we know the radius r or the period T and want to find the other. If these are both known, we can use the equation to calculate the mass of a planet or star.

Watch Physics

Mass and Weight Clarification

This video demonstrates calculations involving Newton’s universal law of gravitation.

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What is the difference between g and G?

g and G are both unchanging constants but have different units.
G is a universal constant that relates force with a pair of masses at a distance, while g relates force with mass and varies with location.
g describes acceleration while G describes gravitational force.
g describes gravitational force while G describes acceleration.

Teacher Support

This video is a thorough demonstration of many of the calculations to be learned in this subsection.

Worked Example

Change in g

The value of g on the planet Mars is 3.71 m/s². If you have a mass of 60.0 kg on Earth, what would be your mass on Mars? What would be your weight on Mars?

Strategy

Weight equals acceleration due to gravity times mass: $W = m g$ . An object’s mass is constant. Call acceleration due to gravity on Mars g_M and weight on Mars W_M.

Solution

Mass on Mars would be the same, 60 kg.

W_{M} = m g_{M} = (60.0 kg) (3.71 {m/s}^{2}) = 223 N

7.4

Discussion

The value of g on any planet depends on the mass of the planet and the distance from its center. If the material below the surface varies from point to point, the value of g will also vary slightly.

Teacher Support

This is a typical mass-weight calculation.

Worked Example

Earth’s g at the Moon

Find the acceleration due to Earth’s gravity at the distance of the moon.

\begin{array}{l} G = 6 .67 \times 10^{–11} {N · m}^{2} {/kg}^{2} \\ Earth-moon distance = 3.84 \times 10^{8} m \\ Earth’s mass = 5.98 \times 10^{24} kg \end{array}

7.5

Express the force of gravity in terms of g.

F = W = m a = m g

7.6

Combine with the equation for universal gravitation.

m g = m G \frac{M}{r^{2}}

7.7

Solution

Cancel m and substitute.

g = G \frac{M}{r^{2}} = (6.67 \times 10^{- 11} \frac{N \cdot m^{2}}{{kg}^{2}}) (\frac{5.98 \times 10^{24} kg}{{(3.84 \times 10^{8} m)}^{2}}) = 2.70 \times 10^{- 3} {m/s}^{2}

7.8

Discussion

The value of g for the moon is 1.62 m/s². Comparing this value to the answer, we see that Earth’s gravitational influence on an object on the moon’s surface would be insignificant.

Teacher Support

[BL][OL] Review the meanings of all the symbols in these equations: F, G, m, M, r, T, and $π$ .

[OL][AL] Have the students memorize the values of G, g, and π to three significant figures.

Practice Problems

6 .

What is the mass of a person who weighs

600\,\text{N}

$6.00\,\text{kg}$
$61.2\,\text{kg}$
$600\,\text{kg}$
$610\,\text{kg}$

7 .

Calculate Earth’s mass given that the acceleration due to gravity at the North Pole is

9.830\text{m/s}^2\!

and the radius of the Earth is

6371\,\text{km}

from pole to center.

$5.94\times10^{17}\,\text{kg}$
$5.94\times10^{24}\text{kg}$
$9.36\times10^{17}\text{kg}$
$9.36\times10^{24}\text{kg}$

Check Your Understanding

8 .

Some of Newton’s predecessors and contemporaries also studied gravity and proposed theories. What important advance did Newton make in the study of gravity that the other scientists had failed to do?

He gave an exact mathematical form for the theory.
He added a correction term to a previously existing formula.
Newton found the value of the universal gravitational constant.
Newton showed that gravitational force is always attractive.

9 .

State the law of universal gravitation in words only.

Gravitational force between two objects is directly proportional to the sum of the squares of their masses and inversely proportional to the square of the distance between them.
Gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
Gravitational force between two objects is directly proportional to the sum of the squares of their masses and inversely proportional to the distance between them.
Gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to the distance between them.

10.

Newton’s law of universal gravitation explains the paths of what?

A charged particle
A ball rolling on a plane surface
A planet moving around the sun
A stone tied to a string and whirled at constant speed in a horizontal circle

Teacher Support

Use the Check Your Answers questions to assess whether students master the learning objectives for this section. If students are struggling with a specific objective, the Check Your Answers will help identify which objective is causing the problem and direct students to the relevant content.