3. A satellite of mass 500 kg orbits the Earth with a period of 6000 s. The Earth has a mass of 5.98 x 1024 kg. (a) Calculate the magnitude of the Earth's gravitational force on the satellite. (b) Determine the altitude of the satellite above the Earth's surface.

To solve this problem, we can use Newton's law of universal gravitation:

F = (G * m1 * m2) / r^2

Where:
F is the gravitational force between two objects,
G is the gravitational constant (6.674 × 10^-11 m^3 kg^-1 s^-2),
m1 is the mass of the first object,
m2 is the mass of the second object, and
r is the distance between the centers of the two objects.

(a) Calculate the magnitude of the Earth's gravitational force on the satellite.

Given:
m1 (mass of the satellite) = 500 kg
m2 (mass of the Earth) = 5.98 x 10^24 kg
r (radius of orbit) = ?

First, we need to find the radius of the satellite's orbit using the period and the gravitational constant.

The formula for the period of a circular orbit is:

T = 2π * sqrt(r^3 / (G * m2))

Rearranging this equation to solve for r:

r^3 = (T^2 * G * m2) / (4π^2)

Now, substitute the known values:

r^3 = (6000 s)^2 * (6.674 × 10^-11 m^3 kg^-1 s^-2) * (5.98 x 10^24 kg) / (4π^2)

Solve for r:

r ≈ 7.09 x 10^6 meters

Now that we have the radius, we can calculate the gravitational force using Newton's law of universal gravitation:

F = (G * m1 * m2) / r^2

F = (6.674 × 10^-11 m^3 kg^-1 s^-2) * (500 kg) * (5.98 x 10^24 kg) / (7.09 x 10^6 meters)^2

F ≈ 3.56 x 10^3 Newtons

Therefore, the magnitude of the Earth's gravitational force on the satellite is approximately 3.56 x 10^3 Newtons.

(b) Determine the altitude of the satellite above the Earth's surface.

The altitude of the satellite above the Earth's surface can be calculated by subtracting the radius of the Earth from the radius of the satellite's orbit:

Altitude = r - Radius of the Earth

Altitude = 7.09 x 10^6 meters - 6.37 x 10^6 meters

Altitude ≈ 0.72 x 10^6 meters

Therefore, the altitude of the satellite above the Earth's surface is approximately 0.72 x 10^6 meters.

To calculate the magnitude of the Earth's gravitational force on the satellite, we need to use Newton's law of universal gravitation:

F = (G * M * m) / r^2

Where:
F is the gravitational force
G is the gravitational constant (6.67430 x 10^-11 m^3 kg^-1 s^-2)
M is the mass of Earth
m is the mass of the satellite
r is the distance between the centers of Earth and the satellite

(a) Calculate the magnitude of the Earth's gravitational force on the satellite:

We are given:
M = 5.98 x 10^24 kg
m = 500 kg

First, let's calculate the radius of the satellite's orbit using the period (T) and the formula for the circumference of a circle:

C = 2 * pi * r

The satellite completes one orbit in 6000 seconds, so the circumference of its orbit is equal to the distance traveled:

C = 2 * pi * r = 6000 seconds

Now, we can solve for r:

r = (6000 s) / (2 * pi) ≈ 955.31 km

Now, substitute the values into the equation for gravitational force:

F = (G * M * m) / r^2
F = (6.67430 x 10^-11 m^3 kg^-1 s^-2) * (5.98 x 10^24 kg) * (500 kg) / (955.31 km)^2

Note: It is necessary to convert the kilometers to meters, as the units in the gravitational constant are given in meters.

F ≈ 3.55 x 10^3 N

Therefore, the magnitude of the Earth's gravitational force on the satellite is approximately 3.55 x 10^3 Newtons.

(b) Determine the altitude of the satellite above the Earth's surface:

The altitude of the satellite above the Earth's surface is equal to the radius of the satellite's orbit minus the radius of the Earth.

Altitude = r - R

Where R is the radius of the Earth (approximately 6.37 x 10^6 meters).

Altitude = (955.31 km - 6.37 x 10^6 m) ≈ 955.31 km

Therefore, the altitude of the satellite above the Earth's surface is approximately 955.31 kilometers.

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