© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Write down Kepler’s law for orbital period
The orbital period $T$ of a satellite around Earth is given by the relation:
$T = 2\pi \sqrt{\frac{r^3}{GM}}$
where:
$r$ is the orbital radius of the satellite.
$M$ is the mass of the Earth.
$G$ is the universal gravitational constant.
Step 2: Express Earth’s mass using period
Rearranging the above equation to solve for $M$, first square both sides:
$T^2 = 4\pi^2 \frac{r^3}{GM}$
Hence,
$M = \frac{4\pi^2}{G} \times \frac{r^3}{T^2}$
Step 3: Apply relative error (uncertainty) rules
The relative uncertainty in $M$, denoted as $\frac{\Delta M}{M}$, is given by summing the respective contributions (with signs depending on exponents) from $r$ and $T$:
$\frac{\Delta M}{M} \;=\; 3\,\frac{\Delta r}{r} \;+\; \bigl(-2\bigr)\,\frac{\Delta T}{T}
\;=\; 3\,\frac{\Delta r}{r} \;-\; 2\,\frac{\Delta T}{T}
$
Since the uncertainty in $r$ is negligible, $\frac{\Delta r}{r} \approx 0$, so:
$\frac{\Delta M}{M} = -\,2\,\frac{\Delta T}{T}
$
We are interested in the magnitude of this relative uncertainty, so:
$\left|\frac{\Delta M}{M}\right| = 2\,\left|\frac{\Delta T}{T}\right|
Step 4: Substitute the given relative uncertainty in $T$
It is given that
$\left|\frac{\Delta T}{T}\right| = 10^{-2}.$
Substituting this, we get:
$\left|\frac{\Delta M}{M}\right| = 2 \times 10^{-2}
Step 5: State the final answer
The relative uncertainty in the mass of the Earth is:
$\boxed{2 \times 10^{-2}}
