iGCSE Physics (0625) 6.1.1 The Earth-Exam Style Questions- New Syllabus
▶️ Answer/Explanation
Detailed solution:
The Earth rotates on its axis from west to east once every $24$ hours.
This rotation causes celestial objects to have an apparent motion in the opposite direction.
Consequently, the Sun appears to rise in the east and set in the west each day.
This daily cycle of day and night is a direct result of this planetary rotation.
Therefore, when viewed from the Earth’s surface, the Sun moves from east to west.
Option A correctly identifies this observed daily directional path.
▶️ Answer/Explanation
Detailed solution:
Average speed is defined as the total distance travelled divided by the total time taken.
For a satellite in a circular orbit, the distance travelled in one complete revolution is the circumference of the circle, given by $2\pi r$.
The time taken for one complete orbit is defined as the orbital period, $T$.
By substituting these into the speed formula $v = \frac{\text{distance}}{\text{time}}$, we get the expression $v = \frac{2\pi r}{T}$.
This matches Option C, which is the standard formula for calculating orbital speed in the syllabus.
▶️ Answer/Explanation
Detailed solution:
The average orbital speed is given by the formula $v = \frac{2\pi r}{T}$, where $T$ is the orbital period.
Rearranging for $T$, we get $T = \frac{2\pi r}{v} = \frac{2 \times \pi \times 7.0 \times 10^{6}}{3500}$.
Calculating this gives $T \approx 12566.37$ seconds.
To convert seconds into minutes, divide by $60$: $12566.37 \div 60 \approx 209.44$ minutes.
To find hours, divide by $60$ again: $209.44 \div 60 \approx 3.49$ hours, which is $3$ hours and $0.49 \times 60 \approx 29$ minutes.
Thus, the orbital period is approximately $3$ hours and $29$ minutes, matching Option D.