Determine the velocity of the Hubble deep space telescope while it is in orbit at a height of 598 km (5.98 x 105 m) above the earth’s surface. The earth’s radius is approximately 6.4 x 106 m. The mass of the earth is 6.0 x 1024 kg.

7.6 x 10^3

To determine the velocity of an object in orbit, we can use the formula for orbital velocity:

V = √(G * M / r)

where V is the velocity, G is the gravitational constant, M is the mass of the planet, and r is the distance from the center of the planet to the object.

First, let's calculate the distance from the center of the Earth to the Hubble deep space telescope. The distance from the Earth's surface to the telescope is given as 598 km. Since the Earth's radius is approximately 6.4 x 10^6 m, the total distance from the center of the Earth to the telescope would be:

r = 6.4 x 10^6 m + 5.98 x 10^5 m
r = 6.99798 x 10^6 m

Now, we can calculate the velocity.

The equation for the orbital velocity of an object is:

V = √(G * M / r)

G is the gravitational constant and is approximately equal to 6.67430 x 10^-11 m^3 kg^-1 s^-2. M is the mass of the Earth and is given as 6.0 x 10^24 kg.

V = √((6.67430 x 10^-11 m^3 kg^-1 s^-2) * (6.0 x 10^24 kg) / (6.99798 x 10^6 m))

Calculating this expression will give us the velocity of the Hubble deep space telescope:

V ≈ 7.71 km/s

Therefore, the velocity of the Hubble deep space telescope while in orbit at a height of 598 km above the Earth's surface is approximately 7.71 km/s.