A satellite of mass 210 kg is launched from a site on Earth's equator into an orbit at 210 km above the surface of Earth.

(a) Assuming a circular orbit, what is the orbital period of this satellite?

(b) What is the satellite's speed in its orbit?

(c) What is the minimum energy necessary to place the satellite in orbit, assuming no air friction?
J

For (a) and (b), the satellite mass will not matter. The orbit distance from the center of the earth is R = 6378 + 210 = 6588 km. Speed is

V = 2 pi R/P
where P is the period.
Get the period from

G M/R^2 = V^2/R

G M = V^2*R = 4 pi^2 R^3/P^2

Solve for P

M is the earth's mass

Energy required = (1/2) m V^2 + GMm[1/Re - (1/(Re + 211,000m)]

m is the satellite mass

There is a small additional term (+ or - that takes into account the rotational speed of the earth at the launch point. That gives the rocket launcher a bit of a boost, especially at the equator.

I need a better explanation than that, and an answer to show you are right. thanks

To determine the orbital period, speed, and minimum energy of the satellite, we can use various formulas and equations in physics. Let's solve each part step by step:

(a) To find the orbital period, we can use Kepler's third law, which states that the square of the orbital period is proportional to the cube of the semi-major axis of the orbit:

T^2 = (4π^2 / GM) * r^3

Where T is the orbital period, G is the gravitational constant (6.67430 × 10^-11 N m^2/kg^2), M is the mass of the Earth (5.972 × 10^24 kg), and r is the radius of the orbit (which is the distance from the center of the Earth to the satellite's altitude, considering Earth's radius as 6371 km).

First, we need to convert the altitude of the satellite to meters:
Altitude = 210 km = 210,000 meters

Now, we can calculate the distance from the center of the Earth:
r = Altitude + Earth's radius = 210,000 + 6,371,000 = 6,581,000 meters

Plugging the values into the formula, we get:
T^2 = (4π^2 / (6.67430 × 10^-11 * 5.972 × 10^24)) * (6,581,000)^3
T^2 ≈ 569,549,854 seconds^2

Taking the square root of both sides, we can find the orbital period:
T ≈ √569,549,854 seconds
T ≈ 23,861 seconds

Therefore, the orbital period of the satellite is approximately 23,861 seconds.

(b) To calculate the satellite's speed in its orbit, we can use the formula:

v = 2πr / T

For this calculation, we already have the value of r from part (a), and we just found T. Plugging the values into the formula, we get:
v = (2π * 6,581,000) / 23,861
v ≈ 4630 meters/second

The satellite's speed in its orbit is approximately 4630 meters/second.

(c) To determine the minimum energy necessary to place the satellite in orbit, we can use the following formula:

E = - (GMm) / (2r)

Where E is the total energy, G is the gravitational constant, M is the Earth's mass, m is the satellite's mass, and r is the radius of the orbit.

Plugging in the given values:
E = - ((6.67430 × 10^-11) * (5.972 × 10^24) * 210) / (2 * 6,581,000)
E ≈ - 1.334 × 10^11 J

Therefore, the minimum energy necessary to place the satellite in orbit, assuming no air friction, is approximately -1.334 × 10^11 Joules. The negative sign indicates that it is the energy required to overcome the gravitational potential.