As Mars orbits the sun in its elliptical orbit, its distance...

As Mars orbits the sun in its elliptical orbit, its distance of closest approach to the center of the sun (at perihelion) is $2067 \times 10^{11} \mathrm{~m},$ and its maximum distance from the center of the sun (at aphelion) is $2.492 \times 10^{11} \mathrm{~m}$. If the orbital speed of Mars at aphelion is $2.198 \times 10^{4} \mathrm{~m} / \mathrm{s},$ what is its orbital speed at perihelion? (You can ignore the influence of the other planets.) 12.7. Consider a spacecraft in an elliptical orbit around the earth. At the low point, or perigee, of its orbit, it is $400 \mathrm{~km}$ above the earth's surface; at the high point, or apogee, it is $4000 \mathrm{~km}$ above the earth's surface. (a) What is the period of the spacecraft's orbit? (b) Using conservation of angular momentum, find the ratio of the spacecraft's speed at perigee to its speed at apogee. (c) Using conservation of energy, find the speed at perigee and the speed at apogee. (d) It is necessary to have the spacecraft escape from the earth completely. If the spacecraft's rockets are fired at perigee, by how much would the speed have to be increased to achieve this? What if the rockets were fired at apogee? Which point in the orbit is more efficient to use?

As Mars orbits the sun in its elliptical orbit, its distance of closest approach to the center of the sun (at perihelion) is $2067 \times 10^{11} \mathrm{~m},$ and its maximum distance from the center of the sun (at aphelion) is $2.492 \times 10^{11} \mathrm{~m}$. If the orbital speed of Mars at aphelion is $2.198 \times 10^{4} \mathrm{~m} / \mathrm{s},$ what is its orbital speed at perihelion? (You can ignore the influence of the other planets.) 12.7. Consider a spacecraft in an elliptical orbit around the earth. At the low point, or perigee, of its orbit, it is $400 \mathrm{~km}$ above the earth's surface; at the high point, or apogee, it is $4000 \mathrm{~km}$ above the earth's surface. (a) What is the period of the spacecraft's orbit?
(b) Using conservation of angular momentum, find the ratio of the spacecraft's speed at perigee to its speed at apogee. (c) Using conservation of energy, find the speed at perigee and the speed at apogee. (d) It is necessary to have the spacecraft escape from the earth completely. If the spacecraft's rockets are fired at perigee, by how much would the speed have to be increased to achieve this? What if the rockets were fired at apogee? Which point in the orbit is more efficient to use?

Key Concepts

Escape Velocity

Escape velocity is the minimum speed an object must have to break free from the gravitational attraction of a celestial body without further propulsion. It is derived from energy conservation principles by equating kinetic energy to the gravitational potential energy needed to reach an infinite distance where the gravitational force becomes negligible. This concept is crucial for mission planning, particularly for transferring spacecraft from bound orbits to trajectories that leave the gravitational influence of the planet.

Conservation of Energy

In a closed system where only conservative forces (like gravity) are acting, the total mechanical energy—which is the sum of kinetic and potential energy—remains constant. This concept allows for the calculation of orbital speeds at different points by relating changes in potential energy to changes in kinetic energy, and it also forms the basis for determining escape conditions.

Kepler's Laws of Planetary Motion

Kepler’s Laws describe the geometric and temporal characteristics of orbits. The third law, in particular, relates the orbital period to the semi-major axis of the ellipse, showing that the square of the period is proportional to the cube of the semi-major axis. This law is fundamental for calculating orbital periods given the sizes and shapes of orbits.

Elliptical Orbits

In orbital mechanics, ellipses are the paths followed by objects under a central inverse-square law force like gravity. Key characteristics include the closest point (periapsis) and the farthest point (apoapsis) from the central body. Understanding the shape and orientation of an ellipse is essential for analyzing orbital motion, determining distances, and relating these distances to orbital speeds and energies.

Conservation of Angular Momentum

For a body moving in a central force field, such as a planet or spacecraft in orbit, angular momentum remains constant in the absence of external torques. This principle implies that the product of the orbital speed and the distance from the center of mass is the same at all points along the orbit, enabling one to relate speeds at different radii (e.g., at periapsis and apoapsis).

Transcript

00:01 So in this question, we are given the fact that mars is orbiting the sun in an elliptical orbit.

00:09 We're given the perihelion and apahelion distances, as well as the orbital speed at epihelion, and asked to calculate the orbital speed at perihelion.

00:20 So basically, this is just going to be a simple conservation of energy application, where we can consider the apahelion state as maybe a final state and the perihelion state as an initial state.

00:38 It really doesn't matter which you consider final and which you consider initial.

00:42 You should end up with the same answer.

00:45 So in that case, the change in potential energy is going to be gmm divided by the distance apocelior.

00:58 Helion distance minus gmm divided by the perihelion distance.

01:05 So this is just uf minus ui.

01:10 And then we're going to do the kinetic energy, the final kinetic energy, so that's going to be one -half m va squared.

01:18 So the speed at apahelion minus one -half m vp squared, so the speed at perihelion.

01:31 And right away, we can do some manipulation of this equation to get rid of the m's because they all do contain the small m.

01:40 So we can go ahead and cross that out.

01:42 And we are trying to isolate for vp.

01:46 So i'm going to take the one half va squared over to the other side.

01:52 So it was a minus, but it's going to become a plus.

01:56 So we've got minus a gmm over rp, sorry, ra.

02:04 Those two negatives become a positive, so gmm or rp, plus one -half va squared.

02:14 Oh, sorry, there shouldn't be an m in here anymore.

02:20 And that will equal one -half vp squared...

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