A satellite of mass M is in a circular orbit of radius R about the centre of the earth. A meteorite

Q: Why do we assume the collision is completely inelastic?

The problem states that the meteorite 'collides with the satellite' and asks for the motion of the 'combined body'. This phrasing implies they stick together, which is the defining characteristic of a completely inelastic collision.

Q: Why does the meteorite only have radial velocity?

The problem describes the meteorite as 'falling towards the earth'. In orbital mechanics, an object simply falling towards the central body without any specified angular motion is assumed to be moving purely radially inwards.

Q: Why isn't the new orbit circular if the energy is still negative?

A negative total energy only tells us the orbit is bound (it won't escape to infinity). However, for an orbit to be perfectly circular, the velocity vector must be exactly perpendicular to the position vector at all times. Since our new velocity has a radial component, the distance to the Earth will change, resulting in an elliptical orbit.

Q: Is kinetic energy conserved in this collision?

No! In any inelastic collision, kinetic energy is lost (usually converted into heat or deformation energy). Only linear momentum is conserved during the brief moment of impact.

Q: What would happen if the total energy turned out to be positive?

If the total mechanical energy was positive, the combined body would have enough energy to overcome the Earth's gravitational pull completely. It would escape to infinity following a hyperbolic trajectory.

Question

A satellite of mass $M$ is in a circular orbit of radius $R$ about the centre of the earth. A meteorite of the same mass falling towards the earth collides with the satellite completely inelastically. The speeds of the satellite and the meteorite are the same just before the collision. The subsequent motion of the combined body will be

Options:

(A) in the same circular orbit of radius $R$

(B) in an elliptical orbit

(C) such that it escapes to infinity

(D) in a circular orbit of a different radius

Sigma Percentile · Accepted Answer

The problem presents a fascinating scenario: a satellite in a stable circular orbit is suddenly struck by a meteorite of equal mass falling directly towards the Earth. To determine the fate of this newly formed combined mass, we must rely on two fundamental pillars of physics: the **conservation of linear momentum** and the **conservation of mechanical energy**.

### Analyzing the Setup

Imagine the satellite, with mass $M$, cruising along its circular orbit of radius $R$. Because it's in a circular orbit, its velocity $\vec{v}_t$ is purely tangential.

Suddenly, a meteorite, also of mass $M$, enters the scene. The problem states it is "falling towards the earth," which means its velocity $\vec{v}_r$ is purely radial (pointing directly towards the Earth's center). We are given that just before the collision, their speeds are identical, so $|\vec{v}_t| = |\vec{v}_r| = v$.

When they collide and stick together, they undergo a **completely inelastic collision**. In such collisions, kinetic energy is not conserved, but linear momentum always is!

### Conservation of Momentum

Let's apply the conservation of linear momentum to find the velocity $\vec{v}'$ of the combined mass immediately after the impact.

The initial momentum is the vector sum of the individual momenta:
$$ \vec{P}_{initial} = M\vec{v}_t + M\vec{v}_r $$

Let's set up a coordinate system where the tangential direction is $\hat{t}$ and the outward radial direction is $\hat{r}$. The satellite's velocity is $v\hat{t}$, and the meteorite's velocity is $-v\hat{r}$ (since it's falling inwards).
$$ \vec{P}_{initial} = Mv\hat{t} - Mv\hat{r} $$

The final momentum belongs to the combined mass $2M$ moving at the new velocity $\vec{v}'$:
$$ \vec{P}_{final} = 2M\vec{v}' $$

Equating the initial and final momenta:
$$ 2M\vec{v}' = Mv\hat{t} - Mv\hat{r} $$
$$ \vec{v}' = \frac{v}{2}\hat{t} - \frac{v}{2}\hat{r} $$

The magnitude of this new velocity is:
$$ |\vec{v}'| = \sqrt{\left(\frac{v}{2}\right)^2 + \left(-\frac{v}{2}\right)^2} = \frac{v}{\sqrt{2}} $$

### Energy and Orbit Classification

To determine the shape of the new orbit, we need to evaluate the total mechanical energy $E_{total}$ of the combined system.

First, let's find the new kinetic energy $K.E.'$:
$$ K.E.' = \frac{1}{2}(2M)(v')^2 = M\left(\frac{v}{\sqrt{2}}\right)^2 = \frac{Mv^2}{2} $$

Recall that the original satellite was in a circular orbit, so its speed $v$ satisfies $v = \sqrt{\frac{GM_e}{R}}$, which means $v^2 = \frac{GM_e}{R}$. Substituting this in:
$$ K.E.' = \frac{GM_e M}{2R} $$

Next, the potential energy $P.E.'$ of the combined mass $2M$ at distance $R$ is:
$$ P.E.' = -\frac{GM_e(2M)}{R} = -\frac{2GM_e M}{R} $$

The total mechanical energy is their sum:
$$ E_{total} = K.E.' + P.E.' = \frac{GM_e M}{2R} - \frac{2GM_e M}{R} = -\frac{3GM_e M}{2R} $$

### The Final Verdict

Since the total mechanical energy is negative ($E_{total} < 0$), the combined mass is gravitationally bound to the Earth. It will not escape to infinity. This leaves us with two possibilities: a circular orbit or an elliptical orbit.

For an orbit to be perfectly circular, the velocity vector must be strictly tangential at all times. However, our new velocity vector $\vec{v}' = \frac{v}{2}\hat{t} - \frac{v}{2}\hat{r}$ clearly has a non-zero radial component ($-\frac{v}{2}\hat{r}$).

Because the mass is moving inwards while also moving sideways, its distance from the Earth will change. A bound orbit with a changing radius is, by definition, an **elliptical orbit**.

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Visualized Solution

$The Collision Setup$

$Initial Velocities$

$Momentum Equation$

$Final Velocity$

$Energy Analysis$

$New Kinetic Energy$

$Total Energy$

$Orbit Classification$

The Sigma Insight: Orbital Motion of a Satellite

Analyzing the Setup

Conservation of Momentum

Energy and Orbit Classification

The Final Verdict

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Visualized Solution

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

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

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

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

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

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

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

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