The mass of the Earth is 5.972kg, and the radius of Earth is 6, 371 km. A 750 kg satellite orbits 35, 800 km above Earth in a perfectly circular orbit.

1. If the gravitational force is the centripetal force keeping the satellite in a stable cicular orbit, how does the centripetal force compare to the gravitational force.
2. If the gravitational force is the centripetal force keeping the satellite in a stable circular orbit, what is the tangential velocity in (m.s) of the satellite's orbit.
3. How long (in hours) does this satellite take to orbit Earth?

To answer these questions, we need to use some basic principles of physics and equations related to circular motion and gravity.

1. The gravitational force between two objects can be calculated using Newton's law of universal gravitation:

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

where F is the gravitational force, G is the gravitational constant (6.67430 × 10^-11 N(m/kg)^2), m1 and m2 are the masses of the two objects, and r is the distance between their centers.

In this case, the satellite is in a stable circular orbit, so the gravitational force acting on it is equal to the centripetal force (F = F_c):

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

In this equation, M is the mass of the Earth, m is the mass of the satellite, v is the tangential velocity of the satellite, and r is the orbital radius.

Rearranging the equation, we can solve for F_c/F:

F_c/F = (G * M * m) / (m * v^2) = (G * M) / v^2

Plugging in the values, we get:

F_c/F = (6.67430 × 10^-11 N(m/kg)^2 * 5.972 × 10^24 kg) / v^2

It is important to note that the centripetal force balances the gravitational force, meaning F_c/F = 1. Therefore, the centripetal force is equal to the gravitational force.

2. To find the tangential velocity (v) of the satellite's orbit, we can rearrange the equation mentioned above:

(G * M * m) / (m * v^2) = 1

(G * M) / v^2 = 1

v^2 = (G * M) / 1

v = √(G * M)

Plugging in the values, we get:

v = √(6.67430 × 10^-11 N(m/kg)^2 * 5.972 × 10^24 kg)

Therefore, the tangential velocity of the satellite's orbit is the square root of the product of the gravitational constant and the mass of the Earth.

3. To find the time it takes for the satellite to orbit Earth, we can use the formula for the period (T) of an orbit:

T = 2πr / v

Plugging in the values, we get:

T = (2π * (r + Earth's radius)) / v

Since the satellite's orbit is 35,800 km above Earth, the orbital radius (r) would be the sum of the Earth's radius (6,371 km) and the altitude (35,800 km).

T = (2π * (6,371 km + 35,800 km)) / v

To convert the result from seconds to hours, divide T by 3600:

T (hours) = (2π * (6,371 km + 35,800 km)) / (3600 * v)

Plugging in the value of v from the previous step, we can find the time it takes for the satellite to orbit Earth in hours.