Calculating Orbital Velocity and Radius Using Energy Conservation

In summary: Regarding the integral sign, I'll keep that in mind for future use. In summary, the original approach taken will not work as the drag force is perpendicular to the Earth's radius. To properly solve this problem, a different approach that takes into account both the kinetic and potential energy must be used. And to make the integral sign larger, it should be placed on its own line.
  • #1
cwill53
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Homework Statement
A satellite with mass 848 kg is in a circular orbit with an orbital speed of 7223 m/s around the earth. After air drag from the earth's upper atmosphere has done ##-1.91\cdot 10^{9} ## J of work on the satellite it will still be in a circular orbit. What are the speed in radius in this new orbit? Hint: You must take account of both kinetic AND potential energy, initially and finally, because both the speed AND the orbit radius change.
Relevant Equations
$$F_{C}=F_{G}$$
##U_{grav}=- \frac{GMm}{r}##
$$\Delta K+\Delta U+\Delta U_{int}=0$$
I know this problem can be solved using energy conservation, but I tried another method that I don't know is correct or not, but yielded a similar result to what my classmates got:
$$F_{C}=F_{G}\Rightarrow \frac{mv^{2}}{r}=\frac{GMm}{r^2}$$
$$\frac{v^2}{r}=\frac{Gm}{r^2}\Rightarrow r=\frac{GM}{v^2}$$
Using the mass of Earth ##M_{E}=5.974\cdot 10^{24}## kg,

##r_{initial}=\frac{\frac{6.674\cdot 10^{-11}m^3}{(kg\cdot s)}(M_{E})}{(7723m/s)^2}=7642161.141m\Rightarrow g## is now ##6.827 m/s^2##

##-1.91\cdot 10^{9} ## J =##\int_{r_{i}}^{r_{f}}\frac{GM_{E}m}{r^2}dr=-GM_{E}m[\frac{1}{r_{f}}-\frac{1}{r_{i}}]##

##-1.91\cdot 10^9J=\frac{GM_{E}m}{r_{f}}+4.424\cdot 10^{10}J##

##\Rightarrow r_{f}=7987281.75m##

##v_{f}=\sqrt{\frac{GM_{e}}{r_{f}}}=7063.1108 m/s##

The answer was about 2% off from the one everyone else got, which was Chegg confirmed to be like 6905 m/s.

By the way, how do you make the integral sign bigger?
 
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  • #2
You seem to have equated the total loss of energy to the loss of just the potential energy. Reread the hint.
 
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cwill53 said:
Homework Statement:: A satellite with mass 848 kg is in a circular orbit with an orbital speed of 7223 m/s around the earth. After air drag from the Earth's upper atmosphere has done ##-1.91\cdot 10^{9} ## J of work on the satellite it will still be in a circular orbit. What are the speed in radius in this new orbit? Hint: You must take account of both kinetic AND potential energy, initially and finally, because both the speed AND the orbit radius change.
Relevant Equations:: $$F_{C}=F_{G}$$
##U_{grav}=- \frac{GMm}{r}##
$$\Delta K+\Delta U+\Delta U_{int}=0$$

I know this problem can be solved using energy conservation, but I tried another method that I don't know is correct or not, but yielded a similar result to what my classmates got:
$$F_{C}=F_{G}\Rightarrow \frac{mv^{2}}{r}=\frac{GMm}{r^2}$$
$$\frac{v^2}{r}=\frac{Gm}{r^2}\Rightarrow r=\frac{GM}{v^2}$$
Using the mass of Earth ##M_{E}=5.974\cdot 10^{24}## kg,

##r_{initial}=\frac{\frac{6.674\cdot 10^{-11}m^3}{(kg\cdot s)}(M_{E})}{(7723m/s)^2}=7642161.141m\Rightarrow g## is now ##6.827 m/s^2##

##-1.91\cdot 10^{9} ## J =##\int_{r_{i}}^{r_{f}}\frac{GM_{E}m}{r^2}dr=-GM_{E}m[\frac{1}{r_{f}}-\frac{1}{r_{i}}]##

##-1.91\cdot 10^9J=\frac{GM_{E}m}{r_{f}}+4.424\cdot 10^{10}J##

##\Rightarrow r_{f}=7987281.75m##

##v_{f}=\sqrt{\frac{GM_{e}}{r_{f}}}=7063.1108 m/s##

The answer was about 2% off from the one everyone else got, which was Chegg confirmed to be like 6905 m/s.

By the way, how do you make the integral sign bigger?
Your approach won't work, as you've done it.

In the relationship between work, force and displacement,

[tex] W = \int_{x_0}^{x_1} \vec F \cdot \vec {dx}, [/tex]

keep in mind that [itex] \vec F [/itex] and [itex] \vec{dx} [/itex] are vectors. The differential work [itex] dW [/itex] is the dot product of the two. In other words, if [itex] dW [/itex] is to increase, [itex] \vec F [/itex] and [itex] \vec{dx} [/itex] need to be more-or-less in the same direction.

But that's not the case in your application, for this particular problem. Here, the Earth's atmosphere's drag force on the satellite is perpendicular to Earth's radius, not parallel to it. So for this problem (as you've treated it),
[tex] \vec F \cdot \vec{dr} = 0. [/tex]
So you'll need to take a different approach.

-- Regarding the size of the integral sign:

In Physics Forums (PF), the smaller integral sign is automatically used when doing inline [itex] \LaTeX [/itex]. If you want to make it larger, put it on it's own line.

Inline notation:
## ... ##
or
[itex] ... [/itex]

To put it on its own line,
$$ ... $$
or
[tex] ... [/tex]
 
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  • #4
collinsmark said:
Your approach won't work, as you've done it.

In the relationship between work, force and displacement,

[tex] W = \int_{x_0}^{x_1} \vec F \cdot \vec {dx}, [/tex]

keep in mind that [itex] \vec F [/itex] and [itex] \vec{dx} [/itex] are vectors. The differential work [itex] dW [/itex] is the dot product of the two. In other words, if [itex] dW [/itex] is to increase, [itex] \vec F [/itex] and [itex] \vec{dx} [/itex] need to be more-or-less in the same direction.

But that's not the case in your application, for this particular problem. Here, the Earth's atmosphere's drag force on the satellite is perpendicular to Earth's radius, not parallel to it. So for this problem (as you've treated it),
[tex] \vec F \cdot \vec{dr} = 0. [/tex]
So you'll need to take a different approach.

-- Regarding the size of the integral sign:

In Physics Forums (PF), the smaller integral sign is automatically used when doing inline [itex] \LaTeX [/itex]. If you want to make it larger, put it on it's own line.

Inline notation:
## ... ##
or
[itex] ... [/itex]

To put it on its own line,
$$ ... $$
or
[tex] ... [/tex]
Oh, now I see where I went wrong. Thanks for the thorough explanation.
 

Related to Calculating Orbital Velocity and Radius Using Energy Conservation

What is a satellite in circular orbit?

A satellite in circular orbit is an object that continuously revolves around a larger object, such as a planet or moon, in a circular path. This orbit is maintained by the gravitational pull of the larger object.

What factors affect the circular orbit of a satellite?

The main factors that affect the circular orbit of a satellite are the mass of the larger object it is orbiting, the distance between the two objects, and the speed of the satellite. These factors determine the strength of the gravitational force and the centripetal force that keeps the satellite in orbit.

What is the difference between a circular and an elliptical orbit?

A circular orbit is a perfectly round path around a larger object, while an elliptical orbit is an oval-shaped path. In a circular orbit, the distance between the satellite and the larger object remains constant, while in an elliptical orbit, the distance varies throughout the orbit.

How does the altitude of a satellite affect its orbit?

The altitude of a satellite, or its distance from the surface of the larger object, determines the speed at which it must travel to maintain a circular orbit. The higher the altitude, the slower the satellite must travel to maintain its orbit.

Can a satellite in circular orbit ever fall back to Earth?

No, a satellite in circular orbit will not fall back to Earth as long as it maintains its speed and altitude. This is because the gravitational pull of the larger object is balanced by the centripetal force of the satellite's motion, keeping it in a stable orbit.

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