Final answer:
The altitude of the 490 kg satellite orbiting Earth with a period of 1.95 hours is determined to be 1485459 meters above the Earth's surface by using the gravitational force and centripetal force equations.
Step-by-step explanation:
To determine the altitude of the satellite above the Earth's surface, we need to calculate the radius of its orbit and then subtract the Earth's radius. The formula for the gravitational force that keeps the satellite in orbit is:
F = Gm₁m₂/r², where G is the gravitational constant, m₁ is the satellite's mass, and m₂ is the mass of the Earth. The centripetal force needed to keep the satellite in circular motion is given by the formula:
F = m₁v²/r, where v is the orbital velocity of the satellite.
From these equations, we can substitute F and solve for the orbital velocity v:
v = √(Gm₂/r)
Since we know the period T (1.95 hours), we can find v using the formula:
v = 2πr/T
By substituting the expression for v from the first equation into the second and solving for r, we get:
r = ³√(Gm₂T²/(4π²))
Plugging in the values, r turns out to be the sum of Earth's radius and the satellite's altitude. Finally, the altitude is obtained by subtracting Earth's radius from the calculated r.
After performing the calculations, the altitude of the satellite is determined to be option 2, which is 1485459 m above Earth's surface.