A satellite is in an elliptic orbit around the earth with aphelion of $6R$ and perihelion of $2R$ where $R = 6400$ km is the radius of the earth. Find eccentricity of the orbit. Find the velocity of the satellite at apogee and perigee. What should be done if this satellite has to be transferred to a circular orbit of radius $6R$ ? [$G = 6.67 \times 10^{-11}$ SI units and $M = 6 \times 10^{24}$ kg]

A satellite is in an elliptic orbit around the earth with aphelion of 6R and perihelion of 2R where R = 6400 k

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Concept understanding — Conservation Of Mechanical Energy

Conservation of Mechanical Energy

The Intuition First

Imagine you're holding a heavy stone at shoulder height. Your arm is tired. That stone has potential energy — energy stored because of its position. Now let it go. As it falls, it speeds up. The potential energy is turning into kinetic energy — the energy of motion. Just before it hits the ground, all the original potential energy has become kinetic energy.

Now imagine the reverse: you throw a ball straight up. It leaves your hand fast (lots of kinetic energy), rises, slows down, stops for an instant at the top (zero kinetic energy), then falls back. At the top, all the kinetic energy you gave it has turned back into potential energy.

This back-and-forth transformation — potential ↔ kinetic — is the heart of the idea. Energy doesn't disappear; it just changes form. That's conservation.

The Precise Statement

Important

Conservation of Mechanical Energy: In an isolated system where only conservative forces (like gravity or an ideal spring) do work, the total mechanical energy of the system remains constant.

Total mechanical energy is the sum of kinetic energy ( $K$ ) and potential energy ( $U$ ):

E_{\text{mech}} = K + U

The law says:

K_{\text{initial}} + U_{\text{initial}} = K_{\text{final}} + U_{\text{final}}

Or, in symbols:

E_{\text{mech, initial}} = E_{\text{mech, final}}

What This Means in Practice

Let's go back to the falling stone. Suppose you hold it 5 metres above the ground. Its mass is 2 kg. Take $g = 10\ \text{m/s}^2$ .

At the top (initial):

$K_i = 0$ (not moving)

$U_i = mgh = 2 \times 10 \times 5 = 100\ \text{J}$

$E_{\text{mech}} = 0 + 100 = 100\ \text{J}$
Just before hitting ground (final):

$U_f = 0$ (height = 0)

$K_f = \frac{1}{2}mv^2$

Conservation says $K_f = 100\ \text{J}$ , so $\frac{1}{2} \times 2 \times v^2 = 100$ , giving $v = 10\ \text{m/s}$ .

You never needed to know the time of fall or acceleration. Energy conservation gave you the speed directly.

The Two Critical Conditions

Watch out

Mechanical energy is not always conserved. It is conserved only when:

No non-conservative forces (like friction, air resistance, or applied pushes/pulls) do work.
The system is isolated — no external forces transfer energy in or out.

If friction is present, some mechanical energy turns into heat (thermal energy). The total energy of the universe is still conserved, but mechanical energy alone is not.

A Simple Example to Cement It …

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The problem involves understanding the properties of elliptic orbits, specifically the relationship between apogee, perigee, semi-major axis, and eccentricity, along with the conservation of mechanical energy and angular momentum to determine velocities. For orbital transfer, we compare the current velocity at a point with the required circular velocity at that same radius.

Concept: Gravitational Potential Energy, Conservation of Mechanical Energy, and Conservation of Angular Momentum in orbital motion.

Eccentricity of the orbit:

The aphelion distance ( $r_a$ ) and perihelion distance ( $r_p$ ) for an elliptic orbit are given by $r_a = a(1+e)$ and $r_p = a(1-e)$ , where $a$ is the semi-major axis and $e$ is the eccentricity.

Given $r_a = 6R$ and $r_p = 2R$ .

Dividing the two equations: $\frac{r_a}{r_p} = \frac{1+e}{1-e} \implies \frac{6R}{2R} = \frac{1+e}{1-e} \implies 3 = \frac{1+e}{1-e}$ .

Solving for $e$ : $3(1-e) = 1+e \implies 3 - 3e = 1+e \implies 2 = 4e \implies e = 0.5$ .
Velocity of the satellite at apogee and perigee:

First, find the semi-major axis $a$ : $r_a + r_p = 2a \implies 6R + 2R = 2a \implies 8R = 2a \implies a = 4R$ .

The velocity $v$ of a satellite in an elliptic orbit at a distance $r$ from the center of the Earth is given by $v = \sqrt{GM \left(\frac{2}{r} - \frac{1}{a}\right)}$ .

Given $G = 6.67 \times 10^{-11}$ SI units, $M = 6 \times 10^{24}$ kg, $R = 6400$ km $= 6.4 \times 10^6$ m.

$GM = (6.67 \times 10^{-11}) \times (6 \times 10^{24}) = 4.002 \times 10^{14} \text{ m}^3/\text{s}^2$ .

At apogee ( $r = r_a = 6R$ ):

$v_a = \sqrt{GM \left(\frac{2}{6R} - \frac{1}{4R}\right)} = \sqrt{GM \left(\frac{1}{3R} - \frac{1}{4R}\right)} = \sqrt{GM \left(\frac{4-3}{12R}\right)} = \sqrt{\frac{GM}{12R}}$

$v_a = \sqrt{\frac{4.002 \times 10^{14}}{12 \times 6.4 \times 10^6}} = \sqrt{\frac{4.002 \times 10^8}{76.8}} \approx 2282.7 \text{ m/s}$ .

At perigee ( $r = r_p = 2R$ ):

$v_p = \sqrt{GM \left(\frac{2}{2R} - \frac{1}{4R}\right)} = \sqrt{GM \left(\frac{1}{R} - \frac{1}{4R}\right)} = \sqrt{GM \left(\frac{4-1}{4R}\right)} = \sqrt{\frac{3GM}{4R}}$

$v_p = \sqrt{\frac{3 \times 4.002 \times 10^{14}}{4 \times 6.4 \times 10^6}} = \sqrt{\frac{12.006 \times 10^8}{25.6}} \approx 6848.3 \text{ m/s}$ .
Transfer to a circular orbit of radius $6R$ : …

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The eccentricity of the elliptical orbit is $\boxed{0.5}$ . The velocity at apogee is $\boxed{2.28 \text{ km/s}}$ and at perigee is $\boxed{6.85 \text{ km/s}}$ . To transfer the satellite to a circular orbit of radius $6R$ , a forward thrust must be applied at apogee to increase its speed by $\boxed{0.95 \text{ km/s}}$ .

When a satellite orbits a central body like the Earth, its path is governed by the gravitational force. This force is conservative, meaning the total mechanical energy (kinetic + potential) of the satellite remains constant in the absence of external forces like air resistance or thrust. Additionally, for central forces, angular momentum is also conserved. These conservation laws are key to understanding orbital mechanics.

An elliptical orbit is characterized by its semi-major axis ( $a$ ) and eccentricity ( $e$ ). The points closest to and farthest from the central body are called perigee and apogee, respectively. Their distances from the center of the Earth are directly related to $a$ and $e$ . The velocity of the satellite varies along the orbit, being highest at perigee and lowest at apogee, consistent with the conservation of angular momentum.

To change an orbit, the satellite's energy must be altered. This is typically done by firing thrusters, which changes the satellite's kinetic energy and thus its total mechanical energy.

Let's break down the problem into three parts:

Given values:

Apogee distance ( $r_a$ ) = $6R$
Perigee distance ( $r_p$ ) = $2R$
Radius of Earth ( $R$ ) = $6400 \text{ km} = 6.4 \times 10^6 \text{ m}$
Gravitational constant ( $G$ ) = $6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2$
Mass of Earth ( $M$ ) = $6 \times 10^{24} \text{ kg}$

First, let's calculate the product $GM$ , which is a fundamental constant for Earth's gravitational field:

$GM = (6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2) \times (6 \times 10^{24} \text{ kg}) = 4.002 \times 10^{14} \text{ m}^3/\text{s}^2$ .

Part 1: Find the eccentricity of the orbit

Relate apogee and perigee distances to semi-major axis and eccentricity: For an elliptical orbit, the distance from the center of the Earth to the apogee ( $r_a$ ) and perigee ( $r_p$ ) are given by:

r_a = a(1+e)

r_p = a(1-e)

where $a$ is the semi-major axis and $e$ is the eccentricity.

2. Substitute the given values:

We are given $r_a = 6R$ and $r_p = 2R$ .

6R = a(1+e) \quad \text{(Equation 1)}

2R = a(1-e) \quad \text{(Equation 2)}

Solve for $a$ and $e$ : Add Equation 1 and Equation 2:

(6R) + (2R) = a(1+e) + a(1-e)

8R = 2a

a = 4R

Now, substitute $a = 4R$ into Equation 1:

6R = (4R)(1+e)

1+e = \frac{6R}{4R} = \frac{3}{2}

e = \frac{3}{2} - 1 = \frac{1}{2}

So, the eccentricity of the orbit is $0.5$.

Part 2: Find the velocity of the satellite at apogee and perigee

Use the Vis-viva equation:

The velocity $v$ of a satellite in an elliptical orbit at any distance $r$ from the central body is given by the Vis-viva equation:
$v^2 = GM \left( \frac{2}{r} - \frac{1}{a} \right)$
Calculate velocities at apogee ( $r_a = 6R$ ) and perigee ( $r_p = 2R$ ):

We have $a = 4R$ .
- At apogee ( $r = r_a = 6R$ ):

v_a^2 = GM \left( \frac{2}{6R} - \frac{1}{4R} \right)

v_a^2 = GM \left( \frac{1}{3R} - \frac{1}{4R} \right)

v_a^2 = GM \left( \frac{4-3}{12R} \right) = \frac{GM}{12R}

v_a = \sqrt{\frac{GM}{12R}}

    Substitute the values: $GM = 4.002 \times 10^{14} \text{ m}^3/\text{s}^2$ and $R = 6.4 \times 10^6 \text{ m}$.

v_a = \sqrt{\frac{4.002 \times 10^{14}}{12 \times 6.4 \times 10^6}} = \sqrt{\frac{4.002 \times 10^{14}}{76.8 \times 10^6}}

v_a = \sqrt{5.211 \times 10^6} \approx 2282.7 \text{ m/s} \approx 2.28 \text{ km/s}

*   **At perigee ($r = r_p = 2R$):**

v_p^2 = GM \left( \frac{2}{2R} - \frac{1}{4R} \right)

v_p^2 = GM \left( \frac{1}{R} - \frac{1}{4R} \right)

v_p^2 = GM \left( \frac{4-1}{4R} \right) = \frac{3GM}{4R}

v_p = \sqrt{\frac{3GM}{4R}}

    Substitute the values:

v_p = \sqrt{\frac{3 \times 4.002 \times 10^{14}}{4 \times 6.4 \times 10^6}} = \sqrt{\frac{12.006 \times 10^{14}}{25.6 \times 10^6}}

    $$v_p = \sqrt{46.90 \times 10^6} \approx 6848.3 \text{ m/s} \approx 6.85 \text{ km/s}$$ …

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Part 1: Find the eccentricity of the orbit

Part 2: Find the velocity of the satellite at apogee and perigee

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