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 formula v = √GM/R+h.

Given:
G = 6.67 × 10^-11 Nm^2kg^2 (Universal Gravitational Constant)
M = 6 × 10^24 kg (mass of Earth)
R = 6.38 × 10^3 km (the orbit of the satellite)
h = 2500 km (height of the satellite above the surface of Earth)

First, we need to convert all the values to SI units:
1 km = 1000 m, so R = 6.38 × 10^3 km = 6.38 × 10^6 m
h = 2500 km = 2500 × 1000 m = 2.5 × 10^6 m

Substituting the values, we get:
v = √(6.67 × 10^-11 Nm^2kg^2 * 6 × 10^24 kg) / (6.38 × 10^6 m + 2.5 × 10^6 m)

Performing the calculations:
v = √(4 × 10^14 Nm^2kg) / (8.88 × 10^6 m)
v = √(4.5 × 10^7) m/s
v ≈ 6708 m/s

Therefore, the orbital velocity of an artificial satellite orbiting Earth at a height of 2500 km is approximately 6708 m/s.