- #1

Pochen Liu

- 52

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- Homework Statement
- A spaceship is launched at 7000 m/s over 3.5km, how much additional speed is needed (via the rocket engine) if the rocket is to orbit Earth at an altitude of 970 km?

- Relevant Equations
- *attached

So using conservation of energy where v0 = 7000 m/s

$$ K_{i} + U_{i} = K_{f} + U_{f} $$

$$\frac{1}{2}mv^{2}_{0} - \frac{GMm}{R} = \frac{1}{2}mv^{2} - \frac{GMm}{r}$$

where R = the radius of the Earth and r = the distance from Earth's center plus the height its orbiting

$$v = \sqrt{7000^{2}-2*6.67*10^{-11}*5.98*10^{24}(\frac{1}{6.37*10^{6}} - \frac{1}{7.34*10^6})}$$

where v is the speed it has when it reaches the velocity when it's at 970km above earth, however to orbit at 970km we need to have a speed of

$$v' = \sqrt{\frac{GM}{r}} = 7371.663102m/s $$

so therefore the extra speed is $$v' - v = 1675 m/s$$ (Depending on how much you round)

I am not given the answer, only told that it's wrong. The answer is acceptable within 5% too.

What am I doing wrong conceptually?

$$ K_{i} + U_{i} = K_{f} + U_{f} $$

$$\frac{1}{2}mv^{2}_{0} - \frac{GMm}{R} = \frac{1}{2}mv^{2} - \frac{GMm}{r}$$

where R = the radius of the Earth and r = the distance from Earth's center plus the height its orbiting

$$v = \sqrt{7000^{2}-2*6.67*10^{-11}*5.98*10^{24}(\frac{1}{6.37*10^{6}} - \frac{1}{7.34*10^6})}$$

where v is the speed it has when it reaches the velocity when it's at 970km above earth, however to orbit at 970km we need to have a speed of

$$v' = \sqrt{\frac{GM}{r}} = 7371.663102m/s $$

so therefore the extra speed is $$v' - v = 1675 m/s$$ (Depending on how much you round)

I am not given the answer, only told that it's wrong. The answer is acceptable within 5% too.

What am I doing wrong conceptually?

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