Orbital Mechanics II
Problems
practice
- The following four statements about circular orbits are equivalent. Derive any one of them from first principles.
- Negative kinetic energy equals half the potential energy (−K = ½U).
- Potential energy equals twice the total energy (U = 2E).
- Total energy equals negative kinetic energy (E = −K).
- Twice the kinetic energy plus the potential energy equals zero (2K + U = 0).
- Determine the minimum energy required to place a large (five metric ton) telecommunications satellite in a geostationary orbit.
- A satellite of mass m is in orbit about the Earth, which has mass M and radius R. (State all answers in terms of the given quantities and fundamental constants.)
- The satellite is initially in an elliptical orbit as shown in the diagram to the right. At perigee (the point of closest approach) the distance from the center of the satellite to the center of the Earth is rp and the speed of the satellite is vp. At apogee (the point when it is furthest from the Earth) the distance from the center of the satellite to the center of the Earth is ra. Determine va, the speed at apogee.
- As the satellite reaches perigee, its speed is changed abruptly so that the satellite enters a circular orbit of radius rp and speed v as shown in the diagram to the right. How much work and what impulse was applied to the satellite to change its orbit?
- Locate the L1, L2, and L3 Lagrange points for the Earth-Sun system using energy considerations. State your answers as distances…
- from the Sun and Earth in meters
- from the Sun as multiples of the Earth's orbital radius (au)
- from the Earth as multiples of the Moon's orbital radius
conceptual
- trajectories-satellite.pdf
The accompanying PDF file shows a satellite in a circular orbit about the Earth. Sketch the new path that the satellite would take if its speed were changed abruptly in the ways described.
numerical
- The planet Mars has two known moons: Phobos and Deimos (fear and panic in Greek). Both are small and orbit quite close to the planet. Phobos is so close that tidal interactions with Mars are dragging it down in its orbit. In 30 million years it will be so close to Mars that tidal forces will shred the tiny moon into a ring. This will occur at a distance 2.4 times the radius of the planet — the Roche limit named after the French astronomer Édouard Roche (1820–1883).
- What is the total mechanical energy of Phobos in its current orbit?
- What total mechanical energy would Phobos have if was orbiting Mars at the Roche limit?
- Where did the lost energy go? Calculate something about Mars that will be different 50 million years in the future using the results of the previous two calculations.
The two moons of Mars Source: NASA/JPL Phobos Deimos mass (× 1015 kg) 10.8 1.8 orbital radius (km) 9,377 23,436 orbital period (Mars days) 0.31891 1.26244 Two planets of the Sun Source: NASA/GSFC Mars Earth mass (× 1024 kg) 0.64171 5.9723 planetary radius (km) 3389.5 6371.0 rotational period (hours) 24.6229 23.9345
algebraic
- Does it take more energy for a satellite in a circular orbit to come to a complete stop and fall to Earth or speed up and escape the Earth?