Question 70 - Review - Chapter 1¶

Problem:¶

The earth has a radius of about $6.37*{10}^{6\;} \;m$.

Question:¶

a) What is the distance from pole to equator over the surface?
b) What is the distance in a straight line from pole to equator (chord)?

Solution:¶

a) The distance from pole to equator along the surface is 1/4 of the circumference of the earth.

$$ \frac{2*\pi *r}{4}=\frac{\pi *r}{2} $$

In [1]:

import numpy as np
r = 6.37e6;
PoolToEq = (np.pi * r) / 2;
print(f'\ta) The distance from pole to equator over the surface = {PoolToEq/1000:g} km.');

	a) The distance from pole to equator over the surface = 10006 km.

b) The chord from pole to equator is the oblique side of the rectangular triangle with rectangular sides = being the radius.

In [2]:

chord = np.sqrt(2*r**2);
print(f'\tb) The chord from pole to equator is {chord/1000:.0f} km.', );

	b) The chord from pole to equator is 9009 km.

سُوۡرَةُ حٰمٓ السجدة / فُصّلَت
ثُمَّ اسْتَوَىٰ إِلَى السَّمَاءِ وَهِيَ دُخَانٌ فَقَالَ لَهَا وَلِلْأَرْضِ ائْتِيَا طَوْعًا أَوْ كَرْهًا قَالَتَا أَتَيْنَا طَائِعِينَ ﴿١١
فَقَضَاهُنَّ سَبْعَ سَمَاوَاتٍ فِي يَوْمَيْنِ وَأَوْحَىٰ فِي كُلِّ سَمَاءٍ أَمْرَهَا ۚ وَزَيَّنَّا السَّمَاءَ الدُّنْيَا بِمَصَابِيحَ وَحِفْظًا ۚ ذَٰلِكَ تَقْدِيرُ الْعَزِيزِ الْعَلِيمِ ﴿١٢

Surah Fussilat: Thereafter turned He to the heaven and it was as smoke, and said Unto it and Unto the earth: come ye twain, willingly or loth. They said: we come willingly. (11) Then He decreed them as seven heavens in two days, and revealed Unto each heaven the command thereof; and We bedecked the nether heaven with lamps and placed therein a guard. That is the ordinance of the Mighty, the Knower. (12)

De hemelen en aarde (waren) een samenhangende massa. Wij hebben ze toen van elkaar gescheiden… Wij hebben de hemel tot een beschermend dak gemaakt … en de dag en de nacht, de zon en de maan geschapen. (Koran 41 11:12)