The potential energy U of an object of mass m that is separated by a distance R from an object of mass M is given by U= -GMm/RA. What is the kinetic energy K of the satellite?Express your answer in terms of the potential energy .
B. Find an expression for the square of the orbital period.

Respuesta :

Answer:

A)  K = ½ U / r , B)     T² = (4π² / G M) r³

Explanation:

A) It asks us for the kinetic energy of the satellite and they give us the relation of the potential synergy

         U = - G M m / r

the force can be found from this expression

          F = - dU / dr

          F = - G M m / r²

The kinetic energy is given by the formula

          K = ½ m v²

let's use Newton's second law

        F = m a

        G M m / r² = m a

where the acceleration is centripetal

        a = v² / r

         G M r² = v² / r

         v² = G M r³

we substitute the expression for the kinetic energy

         K = ½ m v²

         K = ½ m G M r³

in terms of potential energy is

         K = ½ (G M m r²) 1 / r

         K = ½ U / r

B) The period of the orbit

         

We write Newton's second law

            G M m / r² = m a

            a = v² / r

           

in a circular orbit the speed of the velocity is constant

            v = d / t

           

the longitude of the orbit is

             d = 2π r

            v = 2π r / T

we substitute

            G Mm / r² = m (2πr / T)² / r

            G M / r³ = 4π² / T²

            T² = (4π² / G M) r³

(A) The kinetic energy of satellite in terms of potential energy is,

       K =  1/2 U / R.

(B)  The expression for the square of the orbital period is,

         T² = (4π² / G M) R³

Given data:

The potential energy of object is, U.

The mass of object is, m.

The distance from the other object of mass M is, R.

A.

The kinetic energy of the satellite and they give us the relation of the potential synergy,

U = - G M m / R

And the force can be found from this expression

F = - dU / dR

F = - G M m / R²

The kinetic energy is given by the formula

K = 1/2 m v²

Now use Newton's second law

F = m a

G M m / R² = m a

where the acceleration is centripetal

a = v² / R

G M R² = v² / R

v² = G M R³

we substitute the expression for the kinetic energy

K = 1/2 m v²

K = 1/2 m G M R³

in terms of potential energy is

K = 1/2 (G M m R²) 1 / R

 K =  1/2 U / R

Thus, we can conclude that the kinetic energy of satellite in terms of potential energy is K =  1/2 U / R.

(B)

Now we need to obtain the period of the orbit of satellite. For that we   write Newton's second law as

G M m / r² = m a

Since,

a = v² / r

In a circular orbit the speed is constant,

v = d / t

For the longitude of the orbit is

d = 2π R

v = 2π R / T

here,

T is the orbital period of satellite.

 

Substituting the values as,

G Mm / R² = m (2πR / T)² / R

G M / R³ = 4π² / T²

T² = (4π² / G M) R³

Thus, we can conclude that the expression for the square of the orbital period is T² = (4π² / G M) R³.

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