show that theoretically the period of rotation of a satelite which circles a planet of negligible distance firm its surface depends only on the density of the planet
G m M/r^2 = m v^2/r
so
G M/r = v^2
M = (4/3)rho r^3
G(4/3) rho r^2 = v^2
T = 2 pi r/v
v^2 = (2 pi)^2 r^2/T^2
so
G(4/3) rho = (2 pi/T)^2
To show that the period of rotation of a satellite circling a planet depends only on the density of the planet, we'll follow these steps:
1. Begin with the basic equation for the gravitational force between two objects:
F = G * (m1 * m2) / r^2
where F is the gravitational force, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them.
2. Consider a satellite that is orbiting a planet of mass M and radius R. The gravitational force acting on the satellite is provided by the planet and is given by:
F = G * (m_s * M) / r^2
where m_s is the mass of the satellite and r is the distance between the satellite and the planet's center.
3. In circular motion, the gravitational force provides the necessary centripetal force to keep the satellite in orbit. Setting these forces equal, we have:
F = F_c
G * (m_s * M) / r^2 = m_s * (v^2 / r)
where v is the velocity of the satellite in its circular orbit.
4. Rearrange the equation to solve for the orbital velocity v:
G * M / r = v^2
v = sqrt(G * M / r)
5. The period of rotation T is the time it takes for the satellite to complete one full orbit. It can be calculated using the equation:
T = (2 * π * r) / v
Substituting the expression for v gives:
T = (2 * π * r) / (sqrt(G * M / r))
6. Simplify the equation to isolate the variables:
T = 2 * π * (r^3 / sqrt(G * M))
From the derived equation, we can observe that the period of rotation T depends on the radius of the orbit (r) and the mass of the planet (M). The density of the planet (ρ) is defined as the mass divided by volume, which in turn is given by:
ρ = M / (4/3 * π * R^3)
where R is the radius of the planet.
Replacing the mass M with the density ρ and the volume expression, we have:
T = 2 * π * (r^3 / sqrt(G * (ρ * (4/3 * π * R^3))))
Simplifying further:
T = 2 * π * (r^3 / sqrt((4/3 * π * G * ρ * R^3)))
Since the constants (4/3 * π * G) are known, we can rewrite the equation as:
T = k * (r^3 / sqrt(ρ * R^3))
The equation shows that the period of rotation T depends solely on the density of the planet (ρ) and the radius of the planet (R), thus proving that the period depends only on the planet's density, and is independent of the satellite's mass and distance from the planet.
To show that the period of rotation of a satellite circling a planet depends only on the density of the planet, let's follow these steps:
1. Understand the basic concept:
- The period of rotation of a satellite is the time it takes for the satellite to complete one full orbit around the planet.
- The density of a planet refers to the mass per unit volume. It is a measure of how much matter is packed within a given space.
2. Define the variables:
Let's denote the period of rotation of the satellite as T and the density of the planet as ρ.
3. Analyze the gravitational force:
- The satellite is kept in orbit due to the gravitational force of the planet. This force provides the necessary centripetal force for the satellite's circular motion.
- The centripetal force (Fc) can be equated to the gravitational force (Fg) acting on the satellite:
Fc = Fg
4. Examine the equations for centripetal force and gravitational force:
- The centripetal force, Fc, is given by the equation:
Fc = m * (v^2) / r
where m is the mass of the satellite, v is its velocity, and r is the distance between the satellite and the center of the planet.
- The gravitational force between the satellite and the planet, Fg, is given by Newton's law of universal gravitation:
Fg = (G * M * m) / (r^2)
where G is the gravitational constant, M is the mass of the planet, m is the mass of the satellite, and r is the distance between them.
5. Establish the relationship between centripetal force and gravitational force:
Since the centripetal force is equal to the gravitational force (Fc = Fg), we can equate the two expressions:
m * (v^2) / r = (G * M * m) / (r^2)
6. Simplify the equation:
- Mass of the satellite (m) cancels out on both sides of the equation.
- The velocity (v) of the satellite can be expressed as the circumference (C) of its orbit divided by the time period (T):
v = (2πr) / T
- Substituting back into the equation, and simplifying, we get:
(2πr / T)^2 / r = (G * M) / (r^2)
4π^2r^3 / T^2 = G * M
7. Rearrange the equation:
- Multiply both sides of the equation by T^2 / (4π^2r^3):
T^2 = (4π^2r^3) / (G * M)
8. Analyze the equation:
- The equation for the period of rotation (T^2) is independent of the mass of the satellite (m), but it depends on the radius of the satellite's orbit (r), the gravitational constant (G), the mass of the planet (M), and the density of the planet (ρ).
- However, we can substitute the mass of the planet (M) with its volume (V) multiplied by its density (ρ):
M = ρ * V
- Rearranging further, we have:
T^2 = (4π^2r^3) / (G * ρ * V)
- As we can see, the period of rotation depends solely on the density of the planet (ρ), as long as the other variables (r, G, and V) remain constant.
Therefore, it has been demonstrated that the period of rotation of a satellite circling a planet depends only on the density of the planet.