Students will learn to use Newton's Universal Law of Gravity equation to solve problems. They will also learn how the acceleration of gravity on Earth is calculated and a bit about gravitational fields in general.
Key Equations
- \begin{align*}F_G =\frac{Gm_1m_2}{r^2}\end{align*} ; the force of gravity between an object with mass \begin{align*}m_1\end{align*} and another object of mass \begin{align*}m_2\end{align*} and a distance between them of \begin{align*}r\end{align*}.
- \begin{align*}G = 6.67 \times 10^{-11} \ Nm^2 / kg^2\end{align*} ; the universal constant of gravity
- \begin{align*}g = \frac{Gm}{r^2}\end{align*} ; gravitational field strength or gravitational acceleration of a planet with mass \begin{align*}m\end{align*} and radius \begin{align*}r\end{align*}. Note that this is not really a separate equation but comes from Newton’s second law and the law of universal gravitation.
- Some data needed for the problems:
- The radius of Earth is \begin{align*}6.4 \times 10^6 \ m\end{align*}
- The mass of Earth is about \begin{align*}6.0 \times 10^{24} \ kg\end{align*}
- The mass of Sun is about \begin{align*}2.0 \times 10^{30} \ kg\end{align*}
- The Earth-Sun distance is about \begin{align*}1.5 \times 10^{11} \ m\end{align*}
- The Earth-Moon distance is about \begin{align*}3.8 \times 10^8 \ m\end{align*}
- When using the Universal Law of Gravity formula and the constant \begin{align*}G\end{align*} above, make sure to use units of meters and kilograms.
- The direction of the force of gravity is in a straight line between two objects. It is always attractive.
- Newton invented calculus in order to prove that for a spherical object (like Earth) one can assume all of its mass is at the center of the sphere (thus in his formula, one can use the radius of Earth for the distance between a falling rock and Earth).
- Newton's Laws apply to all forces; but when he developed them only one was known: gravity. Newton's major insight --- and one of the greatest in the history of science --- was that the same force that causes objects to fall when released is also responsible for keeping the planets in orbit.
Universal Gravity
Any two objects in the universe, with masses \begin{align*} m_1 \end{align*} and \begin{align*} m_2 \end{align*} with their centers of mass at a distance \begin{align*} r \end{align*} apart will experience a force of mutual attraction along the line joining their centers of mass equal to: \begin{align*} \vec{F_G}=\frac{Gm_1m_2}{r^2} && \text{Universal Gravitation, } \intertext{where G is the Gravitational constant:} G = 6.67300\times10^{-11} \mathrm{m^3 kg^{-1} s^{-2}} \end{align*}
Here is an illustration of this law for two objects, for instance the earth and the sun:
Gravity on the Earth's Surface
On the surface of a planet --- such as earth --- the \begin{align*} r \end{align*} in formula [3] is very close to the radius of the planet, since a planet's center of mass is --- usually --- at its center. It also does not vary by much: for instance, the earth's radius is about 6,000 km, while the heights we consider for this book are on the order of at most a few kilometers --- so we can say that for objects near the surface of the earth, the \begin{align*} r \end{align*} in formula [3] is constant and equal to the earth's radius. This allows us to say that gravity is more or less constant on the surface of the earth. Here's an illustration:
For any object a height \begin{align*} h \end{align*} above the surface of the earth, the force of gravity may be expressed as:
\begin{align*} \vec{F_G} = \frac{Gm_{earth}m_{obj}}{(r_{earth}+h)^2} \intertext{Now we make the approximation that} r_{earth}+h \approx r_{earth} \intertext{then, we can rewrite the force of gravity equation as} \vec{F_G} = \underbrace{\frac{Gm_{earth}}{r_{earth}^2}}_{\vec{g_{earth}}}\times m_{obj} = m_{obj}\times{\vec{g}} \end{align*}
We can do this because the quantity in braces only has constants; we can combine them and call their product \begin{align*} g \end{align*}. Remember, this is an approximation that holds only when the \begin{align*} r \end{align*} in formula [3] is more or less constant.
We call the quantity \begin{align*} mg \end{align*} an object's weight. Unlike an object's mass, an object's weight can change and depends on the gravitational force it experiences. In fact, an object's weight is the magnitude of the gravitational force on it. To find the weight of an object on another planet, star, or moon, use the appropriate values in the formula for the force of gravity.
Simulation
Gravity and Orbits (PhET Simulation)
Time for Practice
- Mo and Jo have been traveling through the galaxy for eons when they arrive at the planet Remulak. Wanting to measure the gravitational field strength of the planet they drop Mo’s lava lamp from the top deck of their spacecraft, collecting the velocity-time data shown below.
velocity (m/s) | time (s) |
---|---|
0 | 0 |
3.4 | 1.0 |
7.0 | 2.0 |
9.8 | 3.0 |
14.0 | 4.0 |
17.1 | 5.0 |
(a) Plot a velocity-time graph using the axes above. Put numbers, labels and units on your axes. Then draw a best-fit line (use a ruler) and use that line to find the gravitational field strength of Remulak. Explain below how you did that.
(b) Mo and Jo go exploring and drop a rock into a deep canyon – it hits the ground in 8.4 s. How deep is the canyon?
(c) If the rock has a mass of 25 g and makes a hole in the ground 1.3 cm deep, what force does the ground exert to bring it to a stop?
(d) Mo and Jo observe the shadows of their lava lamps at different positions on the planet and determine (a la Eratosthenes, the Greek astronomer, around 200 B.C.) that the radius of Remulak is 4500 km. Use that and your result for \begin{align*}g\end{align*} to find the mass of Remulak.
Answer to 5
1a. \begin{align*}\sim 3.4 \ m/s^2\end{align*}
1b. \begin{align*}\sim 120 \ m\end{align*}
1c. \begin{align*}\sim 800 \ N\end{align*}
1d. \begin{align*}\sim 10^{23} \ kg\end{align*}