Extra speed necessary to orbit at 970 km

[Pochen Liu]

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?

 

[BvU]

If the exercise gives you that the rocket engine is going to do work, how can you use conservation of energy ?

 

[Pochen Liu]

If the exercise gives you that the rocket engine is going to do work, how can you use conservation of energy ?

E is conserved as the engine is not on as it leaves at 7000m/s till it reaches 970km. Once it hits 970km the engine will turn on and E is not conserved, however that is not our concern. We only need to determine how much extra speed it needs once its 970km?

If I'm wrong, how else can we go about it?

 

[gneill]

A spaceship is launched at 7000 m/s over 3.5km,...

What does "over 3.5km" mean?

 

[Pochen Liu]

What does "over 3.5km" mean?

It was accelerated over the course of 3.5km to reach 7000 m/s. Which is unrelated? we assume it leaves from the surface of the Earth not 3.5km out

 

[gneill]

It was accelerated over the course of 3.5km to reach 7000 m/s. Which is unrelated? we assume it leaves from the surface of the Earth not 3.5km out

If you are using conservation of energy you can't ignore the altitude at which the velocity is specified.

 

[gneill]

What are the assumptions about the launch conditions? Non-rotating Earth and the spaceship launched vertically? If that's the case, when the ship reaches 970 km it needs to lose its remaining upward velocity and gain the appropriate tangential velocity for the circular orbit. The upward velocity can't be magically transformed into circular velocity.

 

[Janus]

E is conserved as the engine is not on as it leaves at 7000m/s till it reaches 970km. Once it hits 970km the engine will turn on and E is not conserved, however that is not our concern. We only need to determine how much extra speed it needs once its 970km?

If I'm wrong, how else can we go about it?

gneill is pointing you in the right direction. While the answer you get for the speed of the rocket at 970 km seems about right, the velocity vector won't be pointing the right direction when you get to that altitude, so it isn't going just be a matter of increasing speed to match the orbital speed, but matching the orbital velocity.

If you are going to use the conservation of energy approach, you need to go "whole hog", comparing energy to energy, and then work out the velocity difference.

 

[willem2]

I get the same answer. If you launch at the correct angle (use conservation of angular momentum to find this angle), you can get to 970 km altitude with only tangential speed and your calculation is correct. If you launch at a more vertical angle, you will be wasting fuel, and if you launch at a more horizontal angle, you won't get to 970km altitude.
A launch altitude of 3.5 km instead of at the surface is insignificant (if you ignore air friction).
The only other problem I can see, is if your 7000 km/s didn't include the rotation of the earth.