Rishit Vora* [email protected]
The Journal of Irreproducible Results
The experiment demonstrates the approximate derivation of the gravitational force (g), ~100m above the Earth’s surface. Although the gravitational force ($g$) on Earth varies with location, the nominal “average” value on Earth’s surface equates to 9.8 m/s² for all practical intents & purposes [1]. The following experiment uses a miscellanea of crude household equipments & measurement devices to derive the $\text{gravitational force (g)} \approx \text{10.02 m/s}^2$.
An object is released from a measured height of $\text {d = 1.5}\pm\text{0.02m}$. The time ($t$) is measured from the instant of the release to the instant when the object touches the ground. The measurement is carried out via a sophisticated device with a voltage-controlled crystal oscillator (a smartphone). The time is averaged across 10 trials to offset much of the human-induced errors. As this is a Physics experiment, we obviously ignore any air resistance.
$$ \text{Mean t(s) = 0.547} $$
<aside> ⚙ Variables
$$ \begin{aligned} d &= 1.5±0.02m \\ t0 &= 0s \\ u &= 0m \\ t &= 0.547s \\ g &= a = ? \end{aligned} $$
</aside>
We can use the Second Equation of Motion to calculate the acceleration $a$ of the object, displaced by the distance $d$.[2]
$$ \begin{align*} \begin{split} d&=ut+\frac{1}{2}at^2\\1.5&=0(0.547)+\frac{1}{2}a(0.547)^2 \\\therefore \bold a &\bold{\text{ }\approx\text{ }10.02 \text { } m/s^2} \end{split} \end{align*} $$
As the object is displaced through spacetime by no other forces, we can attribute the accelerated displacement to the curvature of spacetime, i.e., Gravity. As derived from the experiment, we can show $\text{gravitational force (g)} \approx \text{10.02 m/s}^2$ (within margin of error of the “actual” value of $g$). A true testament to the replicability of science.