The gravitational field strength at 35700km above earth is .225N/kg. what is the tangental velocity of a 100000kg communication satellite? mass of earth = 5.98E24kg and radius of earth = 6.37E6m

To find the tangential velocity of a communication satellite, we need to use the concept of the gravitational force and centripetal force.

First, let's determine the gravitational force acting on the satellite at an altitude of 35,700 km above the Earth's surface. We can use the formula for gravitational force:

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

Where:
F is the gravitational force
G is the universal gravitational constant (approximately 6.67 x 10^-11 N(m/kg)^2)
m1 is the mass of the satellite
m2 is the mass of the Earth
r is the distance between the satellite and the center of the Earth

Given:
Gravitational field strength at 35,700 km above Earth = 0.225 N/kg
Mass of Earth (m2) = 5.98 x 10^24 kg
Radius of Earth (r) = 6.37 x 10^6 m

The gravitational field strength is given as the force per unit mass, so we can use it to find the gravitational force acting on the satellite. Since the gravitational field strength is equal to the gravitational force per unit mass, we have:

F_gravity = g * m1

Substituting the given value of gravitational field strength (g) and mass of the satellite (m1), we can solve for the gravitational force:

F_gravity = (0.225 N/kg) * (100,000 kg)
F_gravity = 22,500 N

Now, we have the gravitational force acting on the satellite.

Next, we equate this gravitational force to the centripetal force, which is given by:

F_centripetal = (m1 * v^2) / r

Where:
F_centripetal is the centripetal force required to keep the satellite in circular motion
v is the tangential velocity of the satellite
r is the radius of the satellite's orbit (r = Earth's radius + altitude of satellite)

Rearranging this equation, we can solve for the tangential velocity (v):

v = √((F_centripetal * r) / m1)

Substituting the known values, we have:

v = √((22,500 N) * (6.37 x 10^6 m + 35,700,000 m) / (100,000 kg))

Calculating this value, we can now solve for the tangential velocity of the satellite.