Question

This problem concerns the properties of circular orbits for a satellite of mass m orbiting a planet of mass M in an almost circular orbit of radius r.In doing this problem, you are to assume that the planet has an atmosphere that causes a small drag due to air resistance. "Small" means that there is little change during each orbit so that the orbit remains nearly circular, but the radius can change slowly with time.The following questions will ask about the net effects of drag and gravity on the satellite's motion, under the assumption that the satellite's orbit stays nearly circular. Use G if necessary for the universal gravitational constant.U = -(GMm) / (r)K = (GMm) / (2r)Part AThe total mechanical energy of the satellite will __________.increasedecreasestay the samevary in a more complex way than is listed herePart BAs the force of the air resistance acts on the satellite, the radius r of the satellite's orbit will __________.increasedecreasestay the samevary in a more complex way than is listed herePart CAs the force of the air resistance acts on the satellite, the kinetic energy of the satellite will __________.increasedecreasestay the samevary in a more complex way than is listed herePart DAs the force of the air resistance acts on the satellite, the magnitude of the angular momentum of the satellite with respect to the center of the planet will __________.increasedecreasestay the samevary in a more complex way than is listed here

Answer 1

Answer:

Part A: The total mechanical energy of the satellite will increase.

Part B: The radius of the satellite will decrease.

Part C: The kinetic energy of the satellite will increase.

Part D: The angular momentum of the satellite relative to the center of the planet will decrease.

Explanation:

Part A

The total mechanical energy of an object is the sum of its potential energy and kinetic energy.

The question states that for a satellite in a circular orbit around a planet,

• Gravitational potential energy (GPE) = $\displaystyle -\frac{G\cdot M\cdot m}{r}$, and
• Kinetic energy (KE) = $\displaystyle \frac{G\cdot M\cdot m}{2r}$,

where

• $M$ is the mass of the planet,
• $m$ is the mass of the satellite, and
• $r$ is the radius of the orbit.

As a result,

Total mechanical energy = $\displaystyle -\frac{G\cdot M\cdot m}{r} +\frac{G\cdot M\cdot m}{2r} = -\frac{G\cdot M\cdot m}{2r}$.

The satellite will heat up as it moves through the air at high speed. Some of its mechanical energy will be lost as heat. As a result, the total mechanical energy of this satellite will decrease.

Part B

Again, total mechanical energy $\displaystyle =-\frac{G\cdot M\cdot m}{2r}$.

The satellite is trapped in the potential well of the planet. As the negative sign in front of total mechanical energy suggests, the total mechanical energy of the satellite is less than zero. The value of the mechanical energy will become even more negative when the heat is lost to the air. For that to happen, the absolute value of total mechanical energy needs to become larger. The denominator, which is the radius r of the satellite's orbit, needs to become smaller. In other words, the radius of the orbit will decrease as a result of air resistance.

Part C

The question states that the orbit of the satellite is still circular. As a result, the acceleration of the satellite will be the same as its centripetal acceleration.

$\displaystyle a = a_c = \frac{v^{2}}{r}$.

The only force on the satellite is gravitational attraction. The size of the satellite's weight shall be the same as its net force.

$\displaystyle \frac{G\cdot M\cdot m}{r^{2}} = \Sigma F = m\cdot \frac{v^{2}}{r}$.

Rearranging gives:

$\displaystyle v^{2} = G\cdot M\cdot \frac{1}{r}$.

Radius of the orbit $r$ decreases. The value of $1/r$ shall increase. As a result, the velocity of the satellite, $v$, will increase.

Kinetic Energy $\displaystyle = \frac{1}{2} m\cdot v^{2}$.

The kinetic energy of the satellite will also increase as its velocity $v$ increases.

Part D

Workings for part C show that

$\displaystyle v^{2} = G\cdot M\cdot \frac{1}{r}$.

In other words,

$\displaystyle v = \sqrt{G\cdot M}\cdot \sqrt{\frac{1}{r}}$.

The planet is much more massive than the satellite. As a result, the satellite can be treated as a point mass. What will be the angular momentum of the satellite relative to the center of the planet?

$\vec{L} = m \cdot (\vec{v}\times \vec{r})$,

where

• $\vec{L}$ is the angular momentum of the satellite relative to the center of the planet,
• $m$ is the mass of the satellite,
• $\vec{v}$ is the velocity of the satellite, and
• $\vec{r}$ is the position vector of the satellite.

The position vector $\vec{r}$ is a vector that points from the origin (the planet in this case) towards the satellite. The magnitude of that vector is the same as the radius of the orbit, $r$. The satellite is in a circular orbit. As a result, the velocity $\vec{v}$ of the satellite is perpendicular to its position vector at all time. The magnitude of $\vec{v}\times \vec{r}$ will be the same as $v\cdot r$, without vector signs.

Recall that the velocity of the satellite also depends on the radius:

$\displaystyle v = \sqrt{G\cdot M}\cdot \sqrt{\frac{1}{r}}$

Magnitude of angular momentum:

$\abs{\vec{L}}=\displaystyle m \cdot (v\cdot r) = m \cdot \left(\sqrt{G\cdot M}\cdot \sqrt{\frac{1}{r}}\right)\cdot r = (m \cdot\sqrt{G\cdot M})\cdot \sqrt{r}$.

Radius of the satellite becomes smaller. $\sqrt{r}$ becomes smaller. As a result, the magnitude of the angular momentum will decrease.

Note: I referred to the Wikipedia article on angular momentum for the equation $\vec{L} = m \cdot (\vec{v}\times \vec{r})$.