Calculating Orbital Velocity and Radius Using Energy Conservation

[cwill53]

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?

 

[haruspex]

You seem to have equated the total loss of energy to the loss of just the potential energy. Reread the hint.

 

[collinsmark]

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,

W = \int_{x_0}^{x_1} \vec F \cdot \vec {dx},

keep in mind that \vec F and \vec{dx} are vectors. The differential work dW is the dot product of the two. In other words, if dW is to increase, \vec F and \vec{dx} 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),
\vec F \cdot \vec{dr} = 0.
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 \LaTeX. 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]

 

[cwill53]

Your approach won't work, as you've done it.

In the relationship between work, force and displacement,

W = \int_{x_0}^{x_1} \vec F \cdot \vec {dx},

keep in mind that \vec F and \vec{dx} are vectors. The differential work dW is the dot product of the two. In other words, if dW is to increase, \vec F and \vec{dx} 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),
\vec F \cdot \vec{dr} = 0.
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 \LaTeX. 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.