Question

A satellite of mass 2500 kg is orbiting the Earth i
n an elliptical orbit. At the farthest point
from the Earth, its altitude is 3600 km, while at t
he nearest point, it is 1100 km. Calculate
the energy and angular momentum of the satellite an
d its speed at the aphelion and
perihelion.

Answer 1

Answer on Question #61340-Physics-Mechanics-Relativity

A satellite of mass $2500\mathrm{kg}$ is orbiting the Earth in an elliptical orbit. At the farthest point from the Earth, its altitude is $3600\mathrm{km}$ , while at the nearest point, it is $1100\mathrm{km}$ . Calculate the energy and angular momentum of the satellite and its speed at the aphelion and perihelion.

Solution

We know, $U = -\frac{GMm}{r}$ (where G IS gravitational constant, m is satellite's mass, M IS Earth's mass)

At aphelion,

$$r _ {1} = R _ {E A R T H} + 3 6 0 0 k m$$

At perihelion,

$$r _ {2} = R _ {E A R T H} + 1 1 0 0 k m$$

From the conservation of energy:

$$K + U = c o n s t \rightarrow \frac {m v _ {1} ^ {2}}{2} - \frac {G M m}{r _ {1}} = \frac {m v _ {2} ^ {2}}{2} - \frac {G M m}{r _ {2}}$$

From the conservation of angular momentum:

$$m v _ {1} r _ {1} = m v _ {2} r _ {2}$$ $$v _ {2} = \frac {r _ {1}}{r _ {2}} v _ {1}$$ $$\frac {v _ {1} ^ {2}}{2} - \frac {G M}{r _ {1}} = \frac {1}{2} \left(\frac {r _ {1}}{r _ {2}} v _ {1}\right) ^ {2} - \frac {G M}{r _ {2}}$$ $$v _ {1} ^ {2} = G M \frac {\left(\frac {1}{r _ {1}} - \frac {1}{r _ {2}}\right)}{1 - \left(\frac {r _ {1}}{r _ {2}}\right) ^ {2}}$$

1. Speed.

At aphelion,

$$v _ {1} = \sqrt {6. 6 7 3 \cdot 1 0 ^ {- 1 1} \cdot 5 . 9 8 \cdot 1 0 ^ {2 4} \frac {\left(\frac {1}{(6 . 3 7 \cdot 1 0 ^ {6} + 3 . 6 \cdot 1 0 ^ {6})} - \frac {1}{(6 . 3 7 \cdot 1 0 ^ {6} + 1 . 1 \cdot 1 0 ^ {6})}\right)}{1 - \left(\frac {(6 . 3 7 \cdot 1 0 ^ {6} + 3 . 6 \cdot 1 0 ^ {6})}{(6 . 3 7 \cdot 1 0 ^ {6} + 1 . 1 \cdot 1 0 ^ {6})}\right) ^ {2}}} = 4 1 4 0. 5 \frac {m}{s}$$

At perihelion,

$$v _ {2} = \frac {(6 . 3 7 \cdot 1 0 ^ {6} + 3 . 6 \cdot 1 0 ^ {6})}{(6 . 3 7 \cdot 1 0 ^ {6} + 1 . 1 \cdot 1 0 ^ {6})} 4 1 4 0. 5 = 5 5 2 6. 2 \frac {m}{s}$$

2. The energy.

At aphelion,

$$E _ {a} = \frac {1}{2} 2 5 0 0 (4 1 4 0. 5) ^ {2} - \frac {6 . 6 7 3 \cdot 1 0 ^ {- 1 1} \cdot 5 . 9 8 \cdot 1 0 ^ {2 4} \cdot 2 5 0 0}{(6 . 3 7 \cdot 1 0 ^ {6} + 3 . 6 \cdot 1 0 ^ {6})} = - 7. 8 6 \cdot 1 0 ^ {1 0} J.$$

At perihelion,

$$E _ {p} = E _ {a} = - 7. 8 6 \cdot 1 0 ^ {1 0} J.$$

2. The angular momentum.

At aphelion,

$$L _ {a} = 2 5 0 0 (4 1 4 0. 5) (6. 3 7 \cdot 1 0 ^ {6} + 3. 6 \cdot 1 0 ^ {6}) = 1. 0 3 \cdot 1 0 ^ {1 4} \frac {k g m ^ {2}}{s}.$$

At perihelion,

$$L _ {p} = L _ {a} = 1. 0 3 \cdot 1 0 ^ {1 4} \frac {k g m ^ {2}}{s}.$$

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