Essential Physics

Section 2 review

The law of universal gravitation specifies the mutual force of attraction between two objects. The law states that the force of gravity is proportional to the product of the two objects’ masses and inversely proportional to the square of the distance between them. Gravity is the force that keeps satellites and planets in their orbits, which can be circular or elongated (elliptical). If a planet orbits a star, its orbital speed and period depend on the star’s mass and the distance between the two bodies: The planet’s mass is irrelevant! A black hole is an object whose escape velocity exceeds light speed; it can be studied by the effects it has on other objects, such as the radiation emitted by any mass that falls into it.

For more information to research the historical development of the concept of the gravitational force, see

Gravity’s Arc: The Story of Gravity from Aristotle to Einstein and Beyond by D. Darling,
On The Shoulders Of Giants by Stephen J. Hawking (ed.), and
Gravity’s Engines: How Bubble-Blowing Black Holes Rule Galaxies, Stars, and Life in the Cosmos by C. Scharf.

Read the text aloud

law of universal gravitation, satellite, orbit, orbital period, escape velocity, black hole

F = G \frac{m_{1} m_{2}}{r^{2}}

R = \frac{G m}{v^{2}} v = \sqrt{\frac{G m}{R}}

Review problems and questions

Using the formula for the universal law of gravitation, calculate the strength of the gravitational force between two spheres. Each sphere has a mass of 100 kg, and the distance between the centers of the spheres is 2 m. Compare the force to the weight of either sphere. How easy do you think it would be to measure this gravitational force?

Answer: The force of gravity between the two spheres is 1.67×10⁻⁷ N. By contrast, the weight of a 100 kg sphere (on Earth’s surface) is F_w = mg = (9.8)(100)N = 980 N. The gravitational attraction between everyday objects is far, far weaker than people can feel or easily measure.

Asked: force of gravity F between two spheres
Given: sphere mass m = 100 kg, separation distance r = 2 m,
gravitational constant G = 6.67×10⁻¹¹ N m²/kg²
Relationships:

F = \frac{G m_{E} m_{d}}{r^{2}}

Solution:

\begin{array}{l} F & = & G \frac{m^{2}}{r^{2}} \\ = & (6.67 \times 10^{- 11} N m^{2} / {kg}^{2}) \frac{{(100 kg)}^{2}}{{(2 m)}^{2}} = 1.67 \times 10^{- 7} N \end{array}

A 15.0 kg dog rests on Earth’s surface. Since Earth is nearly spherical, assume that you can pretend that Earth’s mass is all at the planet’s center. Earth’s radius is 6,370 km, and its mass is 5.98×10²⁴ kg.
1. What is the strength of the gravitational attraction between the dog and the Earth?
2. Calculate the ratio of this force to the dog’s mass (in kilograms).
3. What is the significance of the ratio you just computed?

Answer:

The Earth and the dog attract one another with a force of 147 N or about 33 lb.
The force divided by the dog’s mass equals 9.8 N/kg.
The ratio just computed is simply the acceleration due to gravity on the Earth’s surface!

Solution:

The force between the Earth and the dog is 147 N.
Asked: gravitational force F between Earth and do
Given: Earth mass m_E = 5.98×10²⁴ kg, dog mass m_d = 15.0 kg, separation r = 6,370 km
Relationship: $F = \frac{G m_{E} m_{d}}{r^{2}}$ Solution: First, convert the distance from kilometers to meters (so that all variables will be in SI units): $6, 370 km \times \frac{1, 000 m}{1 km} = 6.370 \times 10^{6} m$ Next, insert all of the givens and the universal gravitational constant into the relationship formula:
$F = (6.67 \times 10^{- 11} \frac{m^{3}}{{kg s}^{2}}) \frac{5.98 \times 10^{24} kg \times 15 kg}{{(6.37 \times 10^{6} m)}^{2}}$ When faced with an expression like this, you may wish to group base and exponent terms, then add or subtract exponents. Note that dividing by a power of 10, such as 10⁶, is the same as multiplying by the inverse of that power (so you subtract 6 twice to account for the square of 10⁶ in the denominator): $F = (\frac{6.67 \times 5.98 \times 15}{{6.37}^{2}}) [\frac{10^{- 11} \times 10^{24}}{{(10^{6})}^{2}}] N$ Since all of the inputs are in SI units, the force will be in newtons. Simplifying, we get the strength of the gravitational attraction between the Earth and the dog: $\begin{array}{l} F = (14.7) 10^{(- 11 + 24 - 6 - 6)} N \\ F = 146.8 N \end{array}$ Answer: The Earth and the dog attract one another with a force of 147 N, or about 33 lb.
Divide 147 N by 15 kg to get 9.8 N/kg.
The ratio just computed is simply the acceleration due to gravity on the Earth’s surface! Note that g depends on the Earth’s radius as well as its mass. This means that you can make gravity stronger on the Earth’s surface by adding to Earth’s mass or by shrinking the Earth (making it more dense).

Planets Zorg, Kwazu, and Daewok

Planet Zorg has two moons named Kwazu and Daewok. Kwazu and Daewok have the same mass, but Kwazu is three times as far from Zorg as Daewok is.
1. Which moon feels a stronger gravitational pull from Planet Zorg?
2. How many times stronger is the force on that moon when compared to the force on the other (that is, what is the ratio)?


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