The time period of an earth satellite in circular orbit is independent of
(a) the mass of the satell

Q: Why doesn't the satellite's mass matter at all?

Because the gravitational force pulling the satellite in is proportional to its mass, but the inertia (resistance to change in motion) is also proportional to its mass. These two effects perfectly cancel each other out, meaning all objects accelerate at the same rate in a given gravitational field regardless of their mass.

Q: Does the time period depend on the planet's mass?

Yes, absolutely! The mass of the central body (like the Earth, $M_e$) is in the denominator of the time period formula. A more massive planet would pull harder, requiring the satellite to move faster, thus decreasing the time period.

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

The principle remains the same. According to Kepler's Third Law, the square of the time period is proportional to the cube of the semi-major axis of the elliptical orbit. The mass of the orbiting satellite still does not affect the time period.

Q: Does the height of the satellite affect the time period?

Yes. The orbital radius $r$ is the sum of the Earth's radius $R$ and the height $h$ above the surface ($r = R + h$). As height increases, the orbital radius increases, and therefore the time period increases.

Q: Would a heavier satellite fall faster if it lost its orbital speed?

No. Just like Galileo's famous experiment at the Leaning Tower of Pisa, in the absence of air resistance, all objects fall towards the Earth at the exact same acceleration ($g$), regardless of how heavy they are.

Question

The time period of an earth satellite in circular orbit is independent of

Options:

(A) the mass of the satellite

(B) radius of its orbit

(C) both the mass and radius of the orbit

(D) neither the mass of the satellite nor the radius of its orbit

Sigma Percentile · Accepted Answer

### The Physics of Orbital Motion

Imagine you are an engineer tasked with launching two different objects into space: a tiny, lightweight communication satellite and a massive, heavy space station. You want both of them to orbit the Earth at the exact same altitude. A natural question arises: will the heavier space station need to move faster to stay in orbit, or will it take a different amount of time to complete one revolution compared to the tiny satellite?

To answer this, we need to look at the fundamental forces keeping these objects in orbit. For any satellite to move in a circular orbit, it requires a centripetal force. In space, this necessary centripetal force is provided entirely by the gravitational pull of the Earth.

### The Master Equation

Let's set up the equation. The gravitational force $F_g$ between the Earth (mass $M_e$) and the satellite (mass $m$) separated by a distance $r$ is given by Newton's Law of Universal Gravitation:

$$F_g = \frac{G M_e m}{r^2}$$

This force acts as the centripetal force $F_c$, which is required to keep the satellite moving in a circle at a velocity $v$:

$$F_c = \frac{m v^2}{r}$$

Equating the two forces, we get:

$$\frac{G M_e m}{r^2} = \frac{m v^2}{r}$$

Notice what happens here! The mass of the satellite, $m$, appears on both sides of the equation. It elegantly cancels out. This leaves us with the expression for orbital velocity:

$$v = \sqrt{\frac{G M_e}{r}}$$

### The Time Period

The time period $T$ is the time taken to complete one full orbit. Since the satellite travels a distance of $2\pi r$ (the circumference of the orbit) at a speed $v$, we can write:

$$T = \frac{2\pi r}{v}$$

Substituting our expression for $v$ into this equation, we get the final formula for the time period:

$$T = 2\pi \sqrt{\frac{r^3}{G M_e}}$$

### The Grand Conclusion

Look closely at this final formula. The time period $T$ depends only on $r$ (the radius of the orbit) and $M_e$ (the mass of the Earth). The mass of the satellite $m$ is nowhere to be found!

This is a profound realization. It means that in a given gravitational field, the trajectory and the time period of an orbiting body are completely independent of its own mass. Whether it's a feather or a massive asteroid, if they are at the same distance from the Earth, they will orbit it in the exact same amount of time.

Animated Solution for Physics - Gravitation: The time period of an earth satellite in circular orbit is independent of

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Visualized Solution

$Time Period Formula$

$Analyzing Dependencies$

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The Sigma Insight: Orbital Motion of a Satellite

The Physics of Orbital Motion

The Master Equation

The Time Period

The Grand Conclusion

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