The orbital velocity of a satellite is given by v=√GM/R+h, where G= 6.67*10^-11 Nm^2kg^2, is the Universal Gravitational Constant, M=6*10^24 kg, is the mass of Earth, R=6.38*10^3 km, the orbit of the satellite, and h is artificial satellite above the surface of Earth. Use the information to find the orbital velocity of an artificial satellite orbiting Earth at a height of 2500km.

To find the orbital velocity of an artificial satellite orbiting Earth at a height of 2500 km, we can substitute the given values into the equation v = √GM / (R+h).

The given values are:
G = 6.67 * 10^-11 Nm^2kg^2 (Universal Gravitational Constant)
M = 6 * 10^24 kg (mass of Earth)
R = 6.38 * 10^6 m (radius of Earth)
h = 2500 km (height above the surface of Earth)

First, convert the height (h) from kilometers to meters by multiplying by 1000:
h = 2500 km * 1000 = 2.5 * 10^6 m

Now, substitute the values into the equation:

v = √(GM) / (R + h)
= √(6.67 * 10^-11 Nm^2kg^2 * 6 * 10^24 kg) / (6.38 * 10^6 m + 2.5 * 10^6 m)

Simplifying:

v = √(4.002 * 10^14 Nm^2kg^2) / (8.88 * 10^6 m)
= √(4.002 * 10^14) / (8.88 * 10^6) * √(Nm^2kg^2 / m)
≈ 2.99 * 10^4 m/s

Therefore, the orbital velocity of the artificial satellite orbiting Earth at a height of 2500 km is approximately 2.99 * 10^4 m/s.