A spaceship orbits around a planet at a height of 20 km from its surface. Assuming that only gravita

Q: Why do we add the height $h$ to the radius $R$? Can't we just use $h$?

Gravity acts from the center of mass of the planet, not its surface. Therefore, the true distance between the spaceship and the planet's center is the planet's radius plus the height above the surface.

Q: Why did the mass of the spaceship disappear from the equation?

The mass of the spaceship appears on both sides of the equation (in the gravitational force and the centripetal force). It cancels out, proving that orbital velocity depends only on the mass of the planet and the orbital radius, not on the mass of the orbiting object.

Q: Can I use the formula $g = GM/R^2$ to simplify the calculation?

You could find the acceleration due to gravity at the surface and use it, but it adds an unnecessary step. It is much more direct and less prone to rounding errors to use the given values of $G$ and $M$ directly in the velocity formula.

Q: Why did we convert the time period to hours?

The question asks for the number of revolutions in a 24-hour period. To divide the total time by the time period, both must be in the same units. Converting the time period from seconds to hours makes the final division straightforward.

Q: What if the orbit was elliptical instead of circular?

If the orbit were elliptical, the speed would constantly change. We would then have to rely on Kepler's Third Law, which states that the square of the time period is proportional to the cube of the semi-major axis of the ellipse.

Question

A spaceship orbits around a planet at a height of $20	ext{ km}$ from its surface. Assuming that only gravitational field of the planet acts on the spaceship, what will be the number of complete revolutions made by the spaceship in $24	ext{ hours}$ around the planet?
[Take, mass of planet = $8 	imes 10^{22}	ext{ kg}$, radius of planet = $2 	imes 10^6	ext{ m}$, gravitational constant $G = 6.67 	imes 10^{-11}	ext{ N-m}^2/	ext{kg}^2$]

Options:

(A) 11

(B) 17

(C) 13

(D) 9

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine you are an astrophysicist tasked with tracking a spaceship orbiting a distant planet. The planet has a mass $M = 8 	imes 10^{22}	ext{ kg}$ and a radius $R = 2 	imes 10^6	ext{ m}$. The spaceship is cruising at a height $h = 20	ext{ km}$ above the surface.

Before we dive into the complex equations, we must establish the true distance of the spaceship from the center of the planet. Gravity doesn't care about the surface; it acts from the center of mass. Therefore, the orbital radius $r$ is the sum of the planet's radius and the height of the spaceship:

$$r = R + h$$

We must be extremely careful with units here. The height is given in kilometers, while the radius is in meters. Let's convert the height to meters: $h = 20	ext{ km} = 20 	imes 10^3	ext{ m} = 0.02 	imes 10^6	ext{ m}$.

Adding them up, we get the total orbital radius:

$$R + h = 2 	imes 10^6 + 0.02 	imes 10^6 = 2.02 	imes 10^6	ext{ m}$$

### The Master Equation
For the spaceship to maintain a stable circular orbit, it requires a centripetal force pulling it towards the center of the planet. In the vacuum of space, this invisible tether is provided entirely by the gravitational force between the planet and the spaceship.

By equating the gravitational force to the required centripetal force, we get our master equation:

$$F_g = F_c$$

$$\frac{GmM}{(R+h)^2} = \frac{mv^2}{R+h}$$

Notice how the mass of the spaceship, $m$, beautifully cancels out from both sides. This means the orbital velocity is completely independent of how heavy the spaceship is! Solving for the orbital velocity $v$, we get:

$$v = \sqrt{\frac{GM}{R+h}}$$

### Executing the Calculation
Now comes the heavy lifting. We need to substitute the given values into our velocity equation.

$$v = \sqrt{\frac{6.67 	imes 10^{-11} 	imes 8 	imes 10^{22}}{2.02 	imes 10^6}}$$

Handling the powers of ten carefully, the numerator becomes $53.36 	imes 10^{11}$. Dividing this by the denominator gives:

$$v = \sqrt{\frac{53.36 	imes 10^{11}}{2.02 	imes 10^6}} = \sqrt{26.41 	imes 10^5}$$

To make taking the square root easier, we can write this as $\sqrt{2.641 	imes 10^6}$. The square root of $10^6$ is $10^3$, and the square root of $2.641$ is approximately $1.625$. Thus, the orbital velocity is:

$$v \approx 1.625 	imes 10^3	ext{ m/s}$$

### Finding the Time Period
With the spaceship's speed known, we can determine how long it takes to complete one full lap around the planet. The time period $T$ is simply the total distance traveled (the circumference of the orbit) divided by the speed.

$$T = \frac{2\pi(R+h)}{v}$$

Substituting our values:

$$T = \frac{2\pi 	imes 2.02 	imes 10^6}{1.625 	imes 10^3} \approx 7810	ext{ s}$$

Since the question asks for the number of revolutions in $24	ext{ hours}$, it's practical to convert this time period into hours by dividing by $3600$:

$$T \approx \frac{7810}{3600}	ext{ hours} \approx 2.17	ext{ hours}$$

### Final Calculation
Finally, to find the total number of complete revolutions $n$ the spaceship makes in $24	ext{ hours}$, we divide the total time by the time it takes for one revolution:

$$n = \frac{	ext{Total Time}}{T}$$

$$n = \frac{24}{2.17} \approx 11.06$$

Since the question asks for the number of **complete** revolutions, the spaceship completes $11$ full orbits in the given timeframe. This perfectly matches option (a).

Select Answer:

Visualized Solution

$Visualizing the Orbit$

$The Centripetal Force$

$Orbital Velocity$

$Calculating Orbital Radius$

$Calculating Velocity$

$Time Period of Revolution$

$Total Revolutions$

$The Way Forward$

The Sigma Insight: Orbital Motion of a Satellite

Analyzing the Setup Imagine you are an astrophysicist tasked with tracking a spaceship orbiting a distant planet

The Master Equation For the spaceship to maintain a stable circular orbit, it requires a centripetal force pulling it towards the center of the planet

Executing the Calculation Now comes the heavy lifting

Finding the Time Period With the spaceship's speed known, we can determine how long it takes to complete one full lap around the planet

Final Calculation

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List-I

List-II

What is the minimum energy required to launch a satellite of mass $m$ from the surface of a planet of mass $M$ and radius $R$ in a circular orbit at an altitude of $2 R$ ?

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Orbit

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Centripetal Force

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Orbital Velocity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Orbital Radius

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Velocity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Time Period of Revolution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Revolutions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The Way Forward