An 1750 kg weather satellite moves in a circular orbit with a gravitational potential energy of 1.69 x 10^10 J. At it's location, free-fall acceleration is only 6.44 m/s^2. How high above Earth's surface is the satellite?

the radius of the Earth is 6.37E3 km

gravitational force is inversely proportional to the square of the distance

9.81 * (6.37E3)² = 6.44 * (6.37E3 + h)²

To find the height above Earth's surface at which the satellite is orbiting, we can use the formula for gravitational potential energy:

Potential Energy = (mass x acceleration due to gravity x height)

Given:
Mass of satellite (m) = 1750 kg
Gravitational potential Energy (PE) = 1.69 x 10^10 J
Acceleration due to gravity (g) = 6.44 m/s^2

We can rearrange the formula to solve for height (h):

PE = mgh

h = PE / (mg)

Substituting the given values:

h = (1.69 x 10^10 J) / (1750 kg x 6.44 m/s^2)

Calculating the result:

h = 1.69 x 10^10 J / (1750 kg x 6.44 m/s^2)
h ≈ 168690.47 meters

Therefore, the satellite is approximately 168,690.47 meters (or 168.69 kilometers) above Earth's surface.

To determine the height above the Earth's surface at which the satellite is located, we can use the concept of gravitational potential energy and the equation for the potential energy of an object in a circular orbit.

Here's how you can calculate the height above the Earth's surface:

1. Start with the equation for gravitational potential energy:

Potential Energy = mass * gravitational acceleration * height

The given potential energy is 1.69 x 10^10 J.

2. Rearrange the equation to solve for height:

height = Potential Energy / (mass * gravitational acceleration)

3. Substitute the given values into the equation:

height = (1.69 x 10^10 J) / (1750 kg * 6.44 m/s^2)

4. Calculate the height:

height = (1.69 x 10^10 J) / (11290 kg·m/s^2)
height = 1.49 x 10^6 m

Therefore, the satellite is located approximately 1.49 x 10^6 meters above the Earth's surface.

I D K