An object of mass m orbits a star with mass M , where M >> m . In general the bound or bits are elliptical, which means the kinetic energy of the object T and the gravitational potential energy U may vary. For bound orbits, the total energy E is negative. If the orbit is circular, then T and U are both constant.Show that for circular orbits T : U : E = 1 : -2 : -1.

Question

An object of mass m orbits a star with mass M , where M >> m . In general the bound or bits are elliptical, which means the kinetic energy of the object T and the gravitational potential energy U may vary. For bound orbits, the total energy E is negative. If the orbit is circular, then T and U are both constant.Show that for circular orbits T : U : E = 1 : -2 : -1.

Answer 1

well, let's take zero of U at infinity so at radius r from star

U = - G M m/r

the force holding it in orbit is mass * centripetal acceleration
G M m/r^2 = m v^2/r
so
m v^2 = G M m /r Oh, my look at that
so

(1/2)m v^2 = -(1/2) U = Ke

Now the total is the sum of those which is
-(1/2) U + U = (1/2)U

the kinetic energy = (1/2) m v^2

Answer 2

To show that for circular orbits T : U : E = 1 : -2 : -1, we need to express the kinetic energy (T), gravitational potential energy (U), and total energy (E) in terms of the given variables.

Let's start with the kinetic energy (T) of the object in a circular orbit. The kinetic energy is given by the formula:

T = (1/2) * m * v^2

Since the orbit is circular, the velocity (v) can be expressed in terms of the radius (r) and the period (T) of the orbit. The velocity is equal to the circumference of the circular orbit divided by the time taken for one complete revolution:

v = (2 * π * r) / T

Substituting this value of v back into the kinetic energy formula, we get:

T = (1/2) * m * ((2 * π * r) / T)^2 = (1/2) * m * (4 * π^2 * r^2) / T^2

Simplifying, we get:

T = (2 * π^2 * m * r^2) / T^2

Next, let's express the gravitational potential energy (U) of the object in terms of the given variables. The gravitational potential energy is given by the formula:

U = -G * (M * m) / r

Where G is the gravitational constant.

Since the orbit is circular, the radius (r) is constant. Therefore, the gravitational potential energy U is constant.

Now, let's find the total energy (E) of the object in terms of the kinetic energy (T) and gravitational potential energy (U):

E = T + U

Substituting the expressions we derived for T and U, we get:

E = (2 * π^2 * m * r^2) / T^2 - (G * (M * m) / r)

To compare T, U, and E, we can simplify the expression further. Multiply both terms by T^2 to get a common denominator:

E = (2 * π^2 * m * r^2 - G * (M * m) * T^2) / T^2

Since we want to show T : U : E, we can divide all terms by T^2 to get:

T / T^2 : U / T^2 : E / T^2 = (2 * π^2 * m * r^2) / T^2T^2 - (G * (M * m) * T^2) / T^2

Simplifying further:

T / T^2 : U / T^2 : E / T^2 = 2 * π^2 * m * r^2 - G * (M * m)

Now, we can substitute the given mass M >> m. Since M is much larger than m, we can neglect the term 2 * π^2 * m * r^2 compared to G * (M * m):

T / T^2 : U / T^2 : E / T^2 ≈ -G * (M * m)

Finally, since the total energy E is negative for bound orbits, we can express it as:

E / T^2 = -|E| / T^2 = -1

Therefore, T / T^2 : U / T^2 : E / T^2 = 1 : -2 : -1

Hence, we have shown that for circular orbits, T : U : E = 1 : -2 : -1.