Communication satellites are often put in a geo-synchronous orbit, meaning they have an orbital period of 24 hours and stay over the same spot above the equator at all times. Given that the Earth has a mass of 5.97 x 1024 kg and a radius of 6.37 x 106 meters, determine the following: a. What is the orbital height of the satellite relative to the surface of the Earth

h = 35857 km

Explanation:

A geosynchronous orbit can be defined as circular orbit which lies on the Earth's equatorial plane and follows the direction of the Earth's rotation in a period that's equal to the Earth's rotational period and thereby appearing motionless, at a fixed position in the sky relative to the ground observers.

We are given;

Radius of earth(R) = 6.37 x 10^(6) m

Mass of earth (Me) = 5.97 x 10^(24) kg

Gravitational constant = 6.67 × 10^(-11) m³/kg.s²

The earth has a rotational period of 24 hours per day. This gives in seconds

T = 24 × 60 × 60

T = 86400 s

Let's make the height of the orbit from Earth's surface to be h

Also, let ω be the uniform angular velocity in rad/s with which the satellite rotates in the geosynchronous orbit

Now, equating the centripetal force with the gravitational force gives us;

mω²(R + h) = G•Me•m/(R + h)²

m will cancel out. Also ω can be written as 2π/T

Thus,we now have;

(R + h) = ∛(G•Me•T²/(4π²))

Plugging in the relevant values, we have;

(R + h) = ∛(6.67 × 10^(-11) × 5.97 x 10^(24) × 86400²/(4π²))

(R + h) = 42227 Km

Since R = 6.37 x 10^(6)m = 6370 km

Thus;. h = 42227 - 6370 = 35857 km

0.3x0.21x0.08kg=mass 1 sheet

mass 500 = 500xmass 1

volume pile = 0.3x0.2x0.048

density=mass/volume= 500xmass 1/0.3x0.2x0.048

v patient student

explanation:

the masses of the objects and the distance between them