Assuming that the Earth is a sphere of radius 6378 km: (i) Find the length of the equator, correct to the nearest km - Leaving Cert Mathematics - Question b - 2015

The formula for the volume of a sphere is:

$V = \frac{4}{3} \pi r^3$

Plugging in the radius:

$V = \frac{4}{3} \pi (6378)^3$

Calculating:

$V \approx 1.08321 \times 10^{12} \text{ km}^3$

This can be expressed in the required form:

$V = 1.083 \times 10^{12} \text{ km}^3$

So, $a = 1.083$ and $n = 12$.

To find how many times greater the mass of the Sun is compared to the mass of the Earth, we use:

$\text{Ratio} = \frac{\text{Mass of Sun}}{\text{Mass of Earth}} = \frac{1.99 \times 10^{30} \text{ kg}}{5.97 \times 10^{24} \text{ kg}}$

Calculating:

$\text{Ratio} \approx 333.333$

Rounding to the nearest whole number, the mass of the Sun is approximately 333 times greater than the mass of the Earth.