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Step-by-Step Solution
Step 1: Identify the orbital velocity formula
The orbital velocity $V$ of a satellite moving around a planet (assuming a circular orbit) is given by the balance of gravitational force and centripetal force:
$ \frac{m V^2}{r} = \frac{G M m}{r^2} \quad\Rightarrow\quad V = \sqrt{\frac{G M}{r}},$
where:
• $G$ is the universal gravitational constant,
• $M$ is the mass of the planet,
• $r$ is the orbital radius (distance from the center of the planet to the satellite).
Step 2: Determine the orbital radius
The problem states that the spaceship is at a height of $20\,\text{km}$ above the planet’s surface. The planet’s radius is $2\times 10^6\,\text{m}$. Hence:
$ r = R_{\text{planet}} + \text{height} = 2\times 10^6\,\text{m} + 20\times 10^3\,\text{m} = 2.02\times 10^6\,\text{m}.
Step 3: Write the expression for the number of revolutions in the given time
A satellite orbiting at velocity $V$ around a circular orbit of radius $r$ takes a time $T_{\text{orbit}}$ for one revolution given by:
$ T_{\text{orbit}} = \frac{2\pi r}{V}. $
The number of revolutions ($n$) the satellite completes in total time $T$ (24 hours in this case) is:
$ n = \frac{T}{T_{\text{orbit}}} = \frac{T}{\frac{2\pi r}{V}} = \frac{V\,T}{2\pi r}.
Step 4: Substitute the expression for orbital velocity
Since $V = \sqrt{\frac{G M}{r}}$, we have:
$ n = \frac{\sqrt{\frac{G M}{r}}\,T}{2 \pi r} = \sqrt{\frac{G M}{r^3}} \times \frac{T}{2 \pi}.
Step 5: Plug in the numerical values
Given data:
$G = 6.67\times 10^{-11}\,\text{N}\,\text{m}^2/\text{kg}^2$
$M = 8\times 10^{22}\,\text{kg}$
$r = 2.02\times 10^6\,\text{m}$ (as calculated)
$T = 24\,\text{hours} = 24\times 3600\,\text{s}$
Then:
$ n = \sqrt{\frac{6.67\times 10^{-11}\times 8\times 10^{22}}{(2.02\times 10^6)^3}} \times \frac{24\times 3600}{2\pi}.
This expression simplifies numerically to approximately 11.
Step 6: State the final answer
The number of complete revolutions made by the spaceship in 24 hours is 11.
