A satellite is in a circular orbit around the Earth at an altitude of 2.14 106 m.(a) Find the period of the orbit. (Hint: Modify Kepler's third law so it is suitable for objects orbiting the Earth rather than the Sun. The radius of the Earth is 6.38 106 m, and the mass of the Earth is 5.98 1024 kg.) h(b) Find the speed of the satellite. km/s(c) Find the acceleration of the satellite. m/s2 toward the center of the earth

(a) To find the period of the orbit, we first need to find the total distance from the center of the Earth to the satellite. This is the radius of the Earth plus the altitude of the satellite: 6.38 * 10^6 m + 2.14 * 10^6 m = 8.52 * 10^6 m.

We can then use Kepler's third law, which states that the square of the period of a planet's orbit is proportional to the cube of the semi-major axis of its orbit. For objects orbiting the Earth, this can be modified to:

T^2 = 4π^2/GM * r^3

where: T is the period, G is the gravitational constant (6.674 * 10^-11 m^3 kg^-1 s^-2), M is the mass of the Earth (5.98 * 10^24 kg), and r is the distance from the center of the Earth to the satellite.

Substituting the known values:

T^2 = 4π^2/(6.674 * 10^-11 m^3 kg^-1 s^-2 * 5.98 * 10^24 kg) * (8.52 * 10^6 m)^3

Solving for T (the square root of the above equation) will give the period of the orbit in seconds.

(b) The speed of the satellite can be found using the formula for the speed of an object in circular orbit:

v = √(GM/r)

Substituting the known values:

v = √((6.674 * 10^-11 m^3 kg^-1 s^-2 * 5.98 * 10^24 kg) / (8.52 * 10^6 m))

This will give the speed in meters per second. To convert to kilometers per second, divide the result by 1000.

(c) The acceleration of the satellite can be found using the formula for the acceleration of an object in circular motion:

a = v^2 / r

Substitute the value of v from part (b) and r from part (a) to find the acceleration in m/s^2. This acceleration is directed towards the center of the Earth.