A satellite of mass m revolves around the earth of radius R at a height x from its surface. If g is

Q: Why doesn't the mass of the satellite affect its orbital speed?

Because the gravitational force pulling the satellite is proportional to its mass, but the centripetal force required to keep it in orbit is also proportional to its mass. When you equate the two ($F_g = F_c$), the mass $m$ cancels out from both sides.

Q: Why do we use $R+x$ instead of just $x$ for the distance?

Newton's law of gravitation calculates the force based on the distance between the *centers of mass* of the two objects. The distance from the satellite to the Earth's surface is $x$, but the distance to the Earth's center is $R+x$.

Q: What is the difference between $g$ and $G$?

$G$ is the Universal Gravitational Constant, which is the same everywhere in the universe. $g$ is the local acceleration due to gravity, which depends on the mass and radius of the specific planet you are standing on (like Earth).

Q: Can this formula be used for other planets?

Absolutely! You just need to replace $g$ with the surface gravity of that specific planet, and $R$ with the radius of that planet.

Q: What happens if the satellite loses its speed?

If the satellite slows down, the gravitational force will overpower the required centripetal force for that orbit. The satellite will begin to spiral inwards towards the Earth.

Question

A satellite of mass $m$ revolves around the earth of radius $R$ at a height $x$ from its surface. If $g$ is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is

Options:

(A) $gx$

(B) $\frac{gR}{R-x}$

(C) $\frac{gR^2}{R+x}$

(D) $\left(\frac{gR^2}{R+x}\right)^{1/2}$

Sigma Percentile · Accepted Answer

### The Setup: A Satellite in Orbit

Imagine a satellite of mass $m$ gracefully orbiting the Earth. The Earth has a radius $R$, and the satellite is flying at a height $x$ above the surface. This means the total distance from the center of the Earth to the satellite—the orbital radius $r$—is simply $R + x$.

### The Balancing Act: Gravity meets Centripetal Force

For any object to move in a circle, it needs a force pulling it towards the center. This is called the **centripetal force**, given by:

$$F_c = \frac{mv_o^2}{R+x}$$

where $v_o$ is the orbital speed we want to find. But what invisible string is pulling the satellite? It's gravity! According to Newton's Law of Universal Gravitation, the Earth pulls the satellite with a force:

$$F_g = \frac{GM_e m}{(R+x)^2}$$

Since gravity is the only force acting as the centripetal force, we can equate the two:

$$\frac{mv_o^2}{R+x} = \frac{GM_e m}{(R+x)^2}$$

### Simplifying the Equation

Notice how the mass of the satellite, $m$, appears on both sides? It cancels out completely! This is a beautiful fact of orbital mechanics: the speed required to stay in orbit doesn't depend on how heavy the satellite is. A tiny apple and a massive space station would need the exact same speed to stay in the same orbit.

Canceling $m$ and one factor of $(R+x)$ from the denominator, we get:

$$v_o^2 = \frac{GM_e}{R+x}$$

### The Magic of 'g': Connecting to the Surface

Our options don't have the universal gravitational constant $G$ or the mass of the Earth $M_e$. Instead, they use $g$, the acceleration due to gravity on the Earth's surface. We know the relationship:

$$g = \frac{GM_e}{R^2}$$

Rearranging this gives us a handy substitution:

$$GM_e = gR^2$$

### The Final Expression

Let's plug this back into our orbital speed equation:

$$v_o^2 = \frac{gR^2}{R+x}$$

Taking the square root of both sides yields the final orbital speed:

$$v_o = \sqrt{\frac{gR^2}{R+x}} = \left(\frac{gR^2}{R+x}ight)^{1/2}$$

This perfectly matches option (d).

**Bonus Insight:** If the satellite were orbiting extremely close to the Earth's surface (skimming the treetops, ignoring air resistance), $x$ would be almost zero. The formula would simplify to $v_o = \sqrt{gR}$, which is approximately $7.9 	ext{ km/s}$. This is known as the first cosmic velocity!

Animated Solution for Physics - Gravitation: A satellite of mass $m$ revolves around the earth of radius $R$ at a height $x$ from its surface. If $g$ is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is

Select Answer:

Visualized Solution

\text{Visualizing the Orbit}

\text{Gravitational Force}

\text{Centripetal Force}

\text{Equating Forces}

\text{Simplifying the Equation}

\text{Acceleration due to Gravity}

\text{Substituting } GM_e

\text{Final Orbital Speed}

\text{Near-Earth Orbit}

The Sigma Insight: Orbital Motion of a Satellite

The Setup

The Balancing Act

Simplifying the Equation

The Magic of 'g'

The Final Expression

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