- #1

cwill53

- 220

- 40

- 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?

$$F_{C}=F_{G}\Rightarrow \frac{mv^{2}}{r}=\frac{GMm}{r^2}$$

$$\frac{v^2}{r}=\frac{Gm}{r^2}\Rightarrow r=\frac{GM}{v^2}$$

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?

Last edited: