Newton

Newton’s law of gravitation is an inverse square law

$$F_g = \frac {G m_1 m_2}{r^2}$$

where $G = 6.67 \times 10^{-11} \frac {m^3}{kg \cdot s^2}$ is the gravitational constant.

$F_g$ is the gravitational force between point-masses $m_1$ and $m_2$, which are a distance $r$ apart.

For us:

• the two masses are our projectile ($m_p$) and the Earth ($m_e$).
• $r$ is the distance between the centers of our masses. Our projectiles are a negligent distance off of the Earth. Therefore, we can just say that $r = r_e$ (the radius of the Earth). $r$ is basically just a constant for us.

Now, add in the fact that $F = m \, a$ and this tells us that the gravity(acceleration) a projectile feels from the Earth is given by

$$g = \frac {F_g}{m_p} = \frac {G m_e}{(r_e)^2}$$

$$g = \frac {(6.67 \times 10^{-11} \frac {m^3}{kg \cdot s^2}) \cdot (5.972 \times 10^{24} \, kg)}{(6356 \, km)^2} = 9.81 \frac {m}{s^2}$$

A constant!

This is the acceleration we feel on Earth, a downward acceleration, $-9.81 \frac {m}{s^2}$. That means, for most of our investigations, the acceleration due to the Earth can be considered a constant.

$\blacktriangleright$ Gravity is constant acceleration.

The acceleration due to the Earth’s gravity is a constant. It is a constant function with respect to time or position.

$$a(t) = -9.81 \frac {m}{s^2}$$

Acceleration is meters per second PER second. This constant acceleration is a constant rate of change of velocity. Thus, vertical velocity must be a linear function.

$$v(t) = v_0 - 9.81 t \, \frac {m}{s}$$

This velocity is a linear function. The accumulated vertical distance traveled would be the accumulated area under its graph. We have seen that this area (distance) is a described by a quadratic.

$$h(t) = s_0 + v_0 \, t - \frac {9.81}{2} \, t^2$$

• $v_0$ is the initial (vertical) velocity for a projectile thrown into the air.
  
• $s_0$ is the initial height for a projectile thrown into the air.
  
• $h(t)$ is height (vertical distance) as a function of time. Seconds is the domain unit and meters is the range unit.

Projectile Motion

Actually, all of those quantities are vectors.

Both the vertical, $y$, and horizontal, $x$, components of position are functions of time.

We have already seen this equation for the height of a projectile. $$D_y(t) = s_{y_0} + v_{y_0} t - \frac {9.81}{2} t^2$$

Horizontal motion isn’t effected by the Earth’s gravity, since gravity is a downward acceleration. Therefore, the horizontal distance is just affected by the initial horizontal velocity.

$$D_x(t) = s_{x_0} + v_{x_0} t$$

Parabolas

Parabolas are geometric curves. They are defined as the collection of points which are equidistant from a point, called the focus, and a line, called the directrix. In the diagram below, we have rotated our viewpoint so that the directrix is the horizontal line $y = -c$ and the focus is $(0,c)$.

Projectile Trajectory

We have descriptions of the vertical height and the horizontal distance for a projectile under the influence of gravity. Let’s put those together and get height as a function of distance. That will coorespond to our normal experience standing on the ground watching a projectile fly.

• $h_y(t) = s_{y_0} + v_{y_0} t - \frac {9.81}{2} t^2$
• $d_x(t) = s_{x_0} + v_{x_0} t$

First, let’s go back to our separate equations and let’s just assume we are firing the projectile off the ground. Then our initial position will be $0$.

And, let’s use the more common symbol $g$, for the acceleration due to gravity, $g = \frac {9.81}{2} \frac {m^2}{s}$, downward.

• $y = v_{y_0} t - g \, t^2$
• $x = v_{x_0} t$

Solve for $t$ in the horizonal equation.

$$t = \frac {x}{v_{x_0}}$$

Substitute this into the vertical equation.

$$y = v_{y_0} \, \frac {x}{v_{x_0}} - g \left (\frac {x}{v_{x_0}}\right )^2$$

$$y = \frac {v_{y_0}}{v_{x_0}} \, x - \frac {g}{(v_{x_0})^2} \, x^2$$

A quadratic!

The projectile itself follows a parabola trajectory in the air.

Separately...

The projectile’s height is described by a quadratic function in time.

The projectile’s horizontal distance is not affected by gravity, so it is described by a linear function in time.