Module 1.1 Exercises

From 'Fundamentals of Physics (12th ed.) book

Introduction

This is my attempt to understand Physics from the very basics. The first chapter was Measurement and this journal entry will cover the exercises from this chapter.

My math isn’t mathing and I admit that I have a rather poor math skill. As such, I asked Google’s Gemini to review and correct my solutions.

Problem 1

Earth is approximately a sphere and the radius is approximately $ 6.37 \times 10^6 \space m $. What is the (a) circumference in kilometer, (b) surface area in kilometer squared, and (c) volume in cubic kilometer?

My attempt

Circumference of the Earth:

$ C = 2 \pi R $

$ C = (2) \times (3.14) \times (6.37 \times 10^6 \space m) $

$ C = (6.28) \times (6.37 \times 10^6 \space m) $

$ C = 40 \times 10^6 \space m $

$C = 4.0 \times 10^7 \space m$

$C = 4.0 \times 10^7 \space m \left(\frac{1 \space km}{1000 \space m}\right)$

$\boxed{C = 4.0 \times 10^4 \space km}$

Surface Area of the Earth:

$A = 4 \times \pi R^2$

$A = 4 \times 3.14 \times (6.37 \times 10^6 \space m)^2$

$A = 12.56 \times 40.5769 \times 10^8 \space m^2$

$A = 50964586400 \space m^2$

$A = 50964586400 \space m^2 \times \frac{1 \space km}{1000 \space m}$

$A = 50964586,40 \space m \times 1 \space km$

$A = 50964586,40 \space m \times \space km$

$A = 50964586,40 \times km \times \frac{1 \space km}{1000 \space m}$

$A = 50964586,40 \times km \times km$

$\boxed{A = 5,09 \times 10^7 \space km^2}$ (?)

\[V = \frac{4}{3} \times \pi \times R^3 V = \frac{12.56}{3} \times (6.37 \times 10^6m)^3 V = 418.66 \times (6.37 \times 10^6m)^3 V = 418.66 \times 2584748,53 \times 10^14m^3 V = 108213081956,98 \times 10^9m^3 V = 108213081956,98 \times 10^9m^3 \times \frac{1 km}{1000 m} V = 108213081956,98 \times 10^6m^2 \times km V = 108213081956,98 \times 10^6m^2 \times km \times \frac{1 km}{1000 m} \times \frac{1 km}{1000m} V = 108213081956,98 \times 10^(6-6) \times km \times km \times km V = 108213081956,98 km^3 V = 1,08 \times 10^11 km^3 (?)\]