Show that if a satellite orbits very near the surface of a planet with period T, the density (mass per unit volume) of the planet is p(rho)=3pi/(GT)^2
To derive the expression for the density of the planet, we will need to use Newton's law of universal gravitation and the centripetal force acting on the satellite.
Let's start by considering a satellite orbiting very near the surface of a planet. In this scenario, we assume that the satellite is moving in a circular orbit. The gravitational force between the planet and the satellite provides the centripetal force required to keep the satellite in orbit.
According to Newton's law of universal gravitation, the gravitational force between two objects of masses M and m, separated by a distance r, is given by:
F = (G * M * m) / r^2
Where G is the gravitational constant.
Since the satellite is orbiting near the surface of the planet, the distance between the satellite and the planet's center can be approximated as the radius of the planet, denoted as R.
Therefore, the gravitational force acting on the satellite can be expressed as:
F = (G * M * m) / R^2 -- Equation 1
Next, let's consider the centripetal force acting on the satellite. The centripetal force is given by the mass of the satellite multiplied by the acceleration it experiences, which in this case is the centripetal acceleration.
The centripetal force can be expressed as:
F = m * (v^2 / r)
Where v is the velocity of the satellite and r is the distance between the satellite and the planet's center, which is approximated as the radius of the planet, R.
Since the satellite is moving in a circular orbit, the velocity can be expressed as:
v = (2 * pi * R) / T
Where T is the period of the satellite's orbit.
Replacing the velocity in the centripetal force equation, we get:
F = m * ((2 * pi * R) / T)^2 / R
Simplifying further:
F = (4 * pi^2 * m * R) / T^2 -- Equation 2
Remember that the gravitational force acting on the satellite is the same as the centripetal force, as derived in Equation 1. Therefore, we can equate Equations 1 and 2:
(G * M * m) / R^2 = (4 * pi^2 * m * R) / T^2
Simplifying and canceling out similar terms:
G * M / R^3 = 4 * pi^2 / T^2
Now, let's solve for the density of the planet, represented as ρ (rho), which is defined as the mass per unit volume:
ρ = M / (4/3 * pi * R^3)
Substituting the value of M from the equation above:
ρ = (G * M) / (4/3 * pi * R^3) * (R^3 / R^3)
Simplifying further:
ρ = (G * M) / (4 * pi * R^3)
Since we derived that G * M / R^3 = 4 * pi^2 / T^2, we can substitute this in:
ρ = (4 * pi^2 / T^2) / (4 * pi * R^3)
Simplifying further:
ρ = pi / (G * T^2 / R^3)
Finally, substituting the value of G * T^2 / R^3 = (3 * pi) / ρ into the equation, we get:
ρ = 3 * pi / ((G * T^2) / R^3)
Rearranging, we find:
ρ = 3 * pi / (G * T^2 / R^3)
Therefore, the density of the planet is given by:
ρ = 3 * pi / (G * T^2)