Calculate the velocity of a satellite moving in a stable circular orbit about the Earth at a height of 2600 km.

Fgravity=Fcentripetal

GMearth/(radEarth+ 2600km)^2= mv^2/(radearth+2600)

solve for v.

change all distances to meters..

would m be the mass of the earth?

m is mass satellite.

And I forgot one on the left of that line.
Here is is corrected:

GMearth*m/(radEarth+ 2600km)^2= mv^2/(radearth+2600)

does that mean the mass of the statellite cancels? because the mass of the statellite isnt given

yes, the mass of the satellite divides out.

To calculate the velocity of a satellite in a stable circular orbit around the Earth, you can use the formula for the orbital velocity:

\( V = \sqrt{\frac{{G \cdot M}}{{r}}} \)

Where:
- V is the orbital velocity
- G is the gravitational constant (approximately 6.67430 × 10^-11 m^3 kg^-1 s^-2)
- M is the mass of the Earth (approximately 5.972 × 10^24 kg)
- r is the distance between the satellite and the center of the Earth (which is the radius of the Earth plus the height of the satellite above the Earth's surface)

Step 1: Calculate the distance between the satellite and the center of the Earth
Since the height given is above the Earth's surface, we need to add the radius of the Earth (approximately 6,378 km) to it:
\( r = 6,378 km + 2,600 km \)
\( r = 8,978 km \)

Step 2: Convert the distance from kilometers to meters
To use the formula correctly, we need to convert the distance to meters:
\( r = 8,978 km \times 1,000 \)
\( r = 8,978,000 m \)

Step 3: Plug the values into the formula and calculate the velocity
Using the formula:
\( V = \sqrt{\frac{{6.67430 \times 10^{-11} \cdot 5.972 \times 10^{24}}}{{8,978,000}}} \)

Calculating this, we get:
\( V = \sqrt{5.97230 \times 10^{14} \cdot 6.67430 \times 10^{-11}} \)
\( V \approx 7,668.94 \) m/s

Therefore, the velocity of the satellite moving in a stable circular orbit about the Earth at a height of 2600 km is approximately 7,668.94 m/s.