Satellite in orbit losing speed

[Karol]

Homework Statement

A satellite is put into a stationary orbit around earth, i.e. it stays above one point, then it passes through a cloud and looses 10% of it's kinetic energy. it is stabilized again in a new orbit, what's the new radius.

Homework Equations

Attraction force: $F=\frac{GMm}{R^2}$
Acceleration: $a=\frac{V^2}{R}$
Relation of speed with period: $2\pi R=V\cdot T$

The Attempt at a Solution

Finding the initial speed by finding the initial radius:
The initial period is 24 hours, the period is $24\cdot 3600=86,400$
$$\frac{GMm}{R^2}=\frac{V^2}{R}=\frac{4\pi^2 R^2}{T^2}$$
$$R^3=\frac{6.67E-11\cdot 6E24\cdot 86,400^2}{4\pi^2}\rightarrow R=35,904 km$$
The initial speed is:
$$S=VT\rightarrow 42,303,864=V\cdot 86,400\rightarrow V=3076$$
The initial kinetic energy:
$$\frac{1}{2}m\cdot 3076^2=4,257,799$$
The final velocity after deducing 10%:
$$4,257,799\cdot 0.9 \cdot m=\frac{1}{2}m\cdot V^2\rightarrow V^2=8,515,598$$
$$R=\frac{GM}{V^2}=\frac{6.67E-11\cdot 6E24}{8,515,598}=46,996 km$$
It should be R=40,500 km

 

[gneill]

Objects in higher orbits have lower orbital speeds. That's why you found a larger radius for your 10% loss in speed; you didn't account for the fact that to get there you'd have to add gravitational potential energy to reach that orbit.

Have you considered using a conservation of energy approach? There's both KE and PE to deal with.

 

[Karol]

I wanted to delete the thread since i was asked the height above Earth and i forgot to extract the Earth's radius from what i got, so my result is correct

 

[SteamKing]

It would help clarify your calculations if you would indicate units consistently, not every now and then.

 

[haruspex]

I wanted to delete the thread since i was asked the height above Earth and i forgot to extract the Earth's radius from what i got, so my result is correct

Maybe, but gneill is correct that the calculation you posted was wrong. You should take the total energy, PE+KE, subtract 10% of the KE from that, and figure out the radius for the new total.
The question as posted asks for the new radius, not the height above the earth. This suggests that your revised answer, by subtracting Earth's radius, was approximately correct numerically by sheer luck.