A satellite of mass m is launched vertically upwards with an initial speed u from the surface of the

Q: Why is the potential energy at height $R$ taken as $-GMm/2R$ instead of $-GMm/R$?

Gravitational potential energy is measured from the center of the Earth. At a height $R$ above the surface, the total distance from the center is $R + R = 2R$.

Q: Why must the satellite's radial velocity become zero for a circular orbit?

In a perfectly circular orbit, the distance from the center remains constant. If there were any radial velocity, the satellite would move closer to or further from the Earth, making the orbit elliptical.

Q: Can we conserve kinetic energy during the rocket ejection?

No! The ejection is essentially an explosion (an inelastic process in reverse). Internal forces do work, so kinetic energy is not conserved. We must rely strictly on the conservation of momentum.

Q: Why do we conserve momentum separately in the radial and tangential directions?

Momentum is a vector quantity. Since there are no external forces acting on the system *during the instantaneous ejection*, momentum is conserved in all directions independently.

Q: What does the negative sign in the rocket's tangential velocity mean?

The negative sign indicates that the rocket was ejected in the opposite direction to the satellite's new orbital motion. This "backward" ejection provides the forward thrust needed to accelerate the satellite into its orbit.

Question

A satellite of mass $m$ is launched vertically upwards with an initial speed $u$ from the surface of the earth. After it reaches height $R$ ($R =$ radius of the earth), it ejects a rocket of mass $\frac{m}{10}$, so that subsequently the satellite moves in a circular orbit. The kinetic energy of the rocket is ($G$ is the gravitational constant, $M$ is the mass of the earth)

Options:

(A) $\frac{3m}{8} \left( u + \sqrt{\frac{5GM}{6R}} \right)^2$

(B) $\frac{m}{20} \left( u - \sqrt{\frac{2GM}{3R}} \right)^2$

(C) $\frac{m}{20} \left( u^2 + \frac{113}{200} \frac{GM}{R} \right)$

(D) $5m \left( u^2 - \frac{119}{200} \frac{GM}{R} \right)$

Sigma Percentile · Accepted Answer

### The Ascent: Battling Gravity

Imagine you are standing on the surface of the Earth, looking up at a satellite of mass $m$. We launch it vertically upwards with a massive initial speed $u$. As it climbs, Earth's gravity relentlessly pulls it back, converting its kinetic energy into gravitational potential energy.

By the time it reaches a height $h = R$ (which means it is now at a distance of $2R$ from the Earth's center), it has slowed down. To find its exact velocity $v_r$ at this point, we use the principle of conservation of mechanical energy. The total energy at the surface must equal the total energy at height $R$:

$$ \frac{1}{2}mu^2 - \frac{GMm}{R} = \frac{1}{2}mv_r^2 - \frac{GMm}{2R} $$

Solving this for the radial velocity $v_r$, we get:

$$ v_r = \sqrt{u^2 - \frac{GM}{R}} $$

### The Orbital Injection: A Violent Maneuver

Now, the satellite is at the perfect height, but it's moving straight up! To enter a stable circular orbit, two things must happen instantly: its radial velocity must become zero, and it must acquire a precise tangential velocity $v_o$. For an orbit of radius $2R$, this orbital velocity is:

$$ v_o = \sqrt{\frac{GM}{2R}} $$

How does it achieve this? By violently ejecting a rocket of mass $\frac{m}{10}$. Because this is an internal explosion, there are no external forces acting on the system. This means we can rely on our trusty friend: **Conservation of Linear Momentum**. And because momentum is a vector, we must conserve it independently in both the radial and tangential directions.

Let's look at the radial direction first. The initial radial momentum is entirely due to the satellite moving upwards. After the ejection, the satellite's radial momentum must be zero for it to stay in orbit. Therefore, the ejected rocket must carry away *all* of the initial radial momentum:

$$ m v_r = \frac{m}{10} v_r' + \frac{9m}{10} (0) $$

$$ v_r' = 10 v_r = 10 \sqrt{u^2 - \frac{GM}{R}} $$

Next, we look at the tangential direction. Initially, there is zero tangential momentum. After the ejection, the satellite (now with mass $\frac{9m}{10}$) is moving with orbital velocity $v_o$. To keep the total tangential momentum at zero, the rocket must be fired in the exact opposite direction:

$$ 0 = \frac{m}{10} v_t' + \frac{9m}{10} v_o $$

$$ v_t' = -9 v_o = -9 \sqrt{\frac{GM}{2R}} $$

### The Rocket's Fate: Calculating Kinetic Energy

We now know exactly how fast the rocket is moving in both directions. To find its kinetic energy, we simply use the formula $K = \frac{1}{2} m v^2$, where $v^2$ is the sum of the squares of its velocity components:

$$ K_r = \frac{1}{2} \left(\frac{m}{10}\right) (v_r'^2 + v_t'^2) $$

Substituting our velocity components, we get:

$$ K_r = \frac{m}{20} \left[ 100 \left(u^2 - \frac{GM}{R}\right) + 81 \left(\frac{GM}{2R}\right) \right] $$

Now, it's just a matter of careful algebra. Let's expand and group the $\frac{GM}{R}$ terms:

$$ K_r = \frac{m}{20} \left[ 100u^2 - 100\frac{GM}{R} + 40.5\frac{GM}{R} \right] $$

$$ K_r = \frac{m}{20} \left[ 100u^2 - 59.5\frac{GM}{R} \right] $$

To match the options, we factor out $100$ from the bracket:

$$ K_r = \frac{100m}{20} \left[ u^2 - \frac{59.5}{100}\frac{GM}{R} \right] $$

$$ K_r = 5m \left[ u^2 - \frac{119}{200}\frac{GM}{R} \right] $$

And there we have it! A beautiful synthesis of energy and momentum conservation leading us straight to the correct answer.

Select Answer:

Visualized Solution

\text{Trajectory of the Satellite}

\text{Conservation of Mechanical Energy}

\text{Radial Velocity at Height } R

\text{Condition for Circular Orbit}

\text{Conservation of Momentum (Radial)}

\text{Solving for } v_r'

\text{Conservation of Momentum (Tangential)}

\text{Solving for } v_t'

\text{Kinetic Energy of the Rocket}

\text{Simplifying the Expression}

\text{Final Answer}

The Sigma Insight: Orbital Motion of a Satellite

The Ascent

The Orbital Injection

The Rocket's Fate

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