A satellite is moving with a constant speed v in circular orbit around the earth. An object of mass

Q: Why is the final energy at infinity taken as zero?

When an object 'just escapes', it reaches an infinite distance where the gravitational potential energy is zero (since $U \propto 1/r$). Because it 'just' makes it, it has no leftover speed, meaning its kinetic energy is also zero. Thus, the total energy is $0 + 0 = 0$.

Q: Doesn't the object already have kinetic energy before it's ejected?

Yes! While inside the satellite, it is moving at speed $v$, so it has an initial kinetic energy of $\frac{1}{2}mv^2$. The ejection mechanism provides *additional* kinetic energy to bring the total kinetic energy up to $mv^2$. The question asks for the total kinetic energy at the time of ejection, not just the extra energy added.

Q: Why did we substitute $v^2$ for $GM/r$?

The problem gives us the orbital speed $v$ and asks for the answer in terms of $v$. By recalling the formula for orbital velocity, $v = \sqrt{GM/r}$, we can elegantly replace the unknown variables $G$, $M$, and $r$ with the known variable $v$.

Q: Is the ejection velocity the same as the escape velocity?

Not exactly. The escape velocity from that specific orbit is $\sqrt{2GM/r}$, which equals $\sqrt{2}v$. The object must achieve this total speed in the Earth's frame to escape. Its kinetic energy would then be $\frac{1}{2}m(\sqrt{2}v)^2 = mv^2$, which perfectly matches our result!

Q: Does the direction of ejection matter?

For the *total kinetic energy* required to escape, the direction doesn't matter because kinetic energy is a scalar quantity. However, to minimize the *effort* (or energy supplied by the rocket), it's best to eject the object in the forward direction, taking full advantage of the satellite's existing velocity.

Question

A satellite is moving with a constant speed $v$ in circular orbit around the earth. An object of mass '$m$' is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of ejection, the kinetic energy of the object is

Options:

(A) $\frac{3}{2}mv^2$

(B) $2mv^2$

(C) $mv^2$

(D) $\frac{1}{2}mv^2$

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine a satellite orbiting the Earth in a perfectly circular path. It cruises at a constant orbital speed, which we'll call $v$. Inside this satellite is a small object of mass $m$. Because it's inside the satellite, it shares the exact same orbit and the exact same speed $v$.

Now, the mission changes. The object is ejected from the satellite. The goal? To send it so far away that it completely escapes Earth's gravitational pull. But we don't want to waste energy; we want it to *just* escape. This means it should reach infinity, but arrive there completely exhausted, with zero kinetic energy left.

### The Master Equation: Conservation of Energy
Once the object is ejected and is flying through space, the only force acting on it is Earth's gravity. Since gravity is a conservative force, the total mechanical energy of the object remains constant throughout its journey.

This gives us our master equation:
$$ U_i + K_i = U_f + K_f $$

Let's break down the final state first. The object reaches infinity ($r 	o \infty$), so its final potential energy $U_f$ is zero. Because it *just* escapes, its final kinetic energy $K_f$ is also zero.
$$ U_f + K_f = 0 + 0 = 0 $$

### Evaluating the Initial State
Now, let's look at the initial state, right at the moment of ejection. The object is still at the orbital radius $r$. 
Its initial potential energy is:
$$ U_i = -\frac{GMm}{r} $$

This looks a bit messy, but we have a secret weapon. We know the satellite was moving with an orbital speed $v$. The formula for orbital speed is:
$$ v = \sqrt{\frac{GM}{r}} $$

If we square both sides, we get:
$$ v^2 = \frac{GM}{r} $$

Look closely at our potential energy equation. We can substitute $\frac{GM}{r}$ with $v^2$!
$$ U_i = -m \left( \frac{GM}{r} ight) = -mv^2 $$

### Final Calculation
We are ready to solve for the initial kinetic energy, $K_i$. Let's plug everything back into our conservation of energy equation:
$$ -mv^2 + K_i = 0 $$

Moving the potential energy term to the other side, we get our final answer:
$$ K_i = mv^2 $$

**The kinetic energy of the object at the time of ejection must be exactly $mv^2$.**

It's important to note that this is the total kinetic energy of the object in the Earth's frame of reference. If you were asked for the energy *supplied* by the ejection mechanism, you would have to subtract the kinetic energy it already had while sitting inside the satellite ($\frac{1}{2}mv^2$). But the question simply asks for the kinetic energy *of the object* at the time of ejection, which is $mv^2$.

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Visualized Solution

$Visualizing the Orbit$

$The Escape Condition$

$Conservation of Mechanical Energy$

$Initial Potential Energy$

$Solving for Initial Kinetic Energy$

$Conclusion$

The Sigma Insight: Escape Speed and Motion of Satellites

Analyzing the Setup Imagine a satellite orbiting the Earth in a perfectly circular path

The Master Equation

Evaluating the Initial State Now, let's look at the initial state, right at the moment of ejection

Final Calculation We are ready to solve for the initial kinetic energy, $K_{i}$

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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 Escape Condition

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conservation of Mechanical Energy

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for Initial Kinetic Energy

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