As the satellite moves to a higher orbit, its gravitational potential energy increases due to the increase in distance from the Earth's center.

If the speed of a satellite is doubled, the orbital radius will decrease by a factor of four, as orbital speed is inversely proportional to the square root of the radius.

Increasing the speed of a satellite in a circular orbit will cause its orbit to become elliptical.

The gravitational force is inversely proportional to the square of the distance; halving the radius increases the force by a factor of four.

Increasing the speed of a satellite in a circular orbit will cause it to move into an elliptical orbit.

The orbital speed can be calculated using the formula v = √(GM/r). For a height of 300 km, the speed is approximately 7.9 km/s.

A polar orbit allows the satellite to pass over the entire surface of the Earth as the planet rotates beneath it.

For a satellite in a circular orbit, the kinetic energy is less than the potential energy, as K.E. = -1/2 P.E.

Low Earth orbit satellites typically operate at altitudes ranging from about 200 km to 2000 km above the Earth's surface.

For a satellite in a stable orbit, the centripetal force required for circular motion equals the gravitational force acting on the satellite.

A geostationary satellite has an orbital period equal to the Earth's rotation period, which is 24 hours.

In a stable orbit, the net force acting on the satellite is zero because the gravitational force provides the necessary centripetal force.

The orbital period of a satellite increases when it is moved to a higher orbit, according to Kepler's third law.

For a satellite in a stable circular orbit, the centripetal acceleration is equal to the gravitational acceleration.

A satellite in a circular orbit possesses both kinetic energy due to its motion and potential energy due to its position in the gravitational field.

If a satellite's altitude is doubled, its orbital speed decreases by √2.

As the altitude increases, the orbital period increases due to the greater distance from the Earth's center.

If a satellite's speed is less than the required orbital speed, it will not have enough centripetal force to maintain its orbit and will fall back to Earth.

The gravitational force acting on the satellite will double if its mass is doubled, as gravitational force is directly proportional to mass.

The gravitational force F acting on a satellite at height h is given by F = GmM/(R+h)^2, where G is the gravitational constant.

The distance from the center of the Earth to a geostationary satellite is approximately 42000 km.

The orbital speed v of a satellite is given by v = sqrt(GM/(R+h)), where M is the mass of the Earth and G is the gravitational constant.

A geostationary satellite orbits at a height of approximately 36,000 km above the Earth's surface, which is about R (the radius of the Earth) plus the height of the satellite.

The gravitational force acting on the satellite is given by Newton's law of gravitation, which states that F = G * (M * m) / (R + h)^2, where M is the mass of the Earth and m is the mass of the satellite.

At a height R above the Earth's surface, the gravitational acceleration is g/4, derived from the formula g' = g(R/(R+h))^2.

The radius of a geostationary orbit is approximately 3R, where R is the radius of the Earth.

The gravitational force is inversely proportional to the square of the distance. If the radius doubles, the force decreases to one-fourth.

The orbital speed v is given by v = √(GM/r). If r is doubled, v decreases by a factor of √2.

The gravitational force acting on a satellite depends on the mass of the satellite, the mass of the Earth, and the distance from the Earth according to Newton's law of gravitation.

The time period of a satellite in a low Earth orbit is approximately 90 minutes due to its close proximity to the Earth.