Question

1.Consider a satellite of the earth in a circular orbit of radius R. Let M and m be the mass of the earth and that of the satellite, respectively. Show that the centripetal acceleration of the satellite is a_R = -(v^2/R)*(r/r) where v = |v| is the magnitude of the velocity V and r/r is a unit vector in the radial direction.

2.Using Newton's second low of motion and the law of universal gravitation, determine the speed v=|V| and the angular speed w for the circular orbit.

3.For two satellites in two different circular orbits with radii R1 and R2, respectively, prove kepler's second and third laws for this special case.

Answer 1

1)

The position of an object in a plane can be converted from polar to cartesian coordinates through the equations

Expressing ? as a function of time gives equations for the cartesian coordinates as a function of time in uniform circular motion:

Differentiation with respect to time gives the components of the velocity vector:

Velocity in circular motion is a vector tangential to the trajectory of the object. Furthermore, even though the speed is constant the velocity vector changes direction over time. Further differentiation leads to the components of the acceleration (which are just the rate of change of the velocity components):

The acceleration vector is perpendicular to the velocity and oriented towards the centre of the circular trajectory. For that reason, acceleration in circular motion is referred to as centripetal acceleration.

The absolute value of centripetal acceleration may be readily obtained by

2)

a= v^2/R

==> from newtons 2nd law

F=ma

==> F=mv^2/R

==> from law of gravitation

Force = GMm/R^2

==> GMm/R^2=mv^2/R

==> v= sqrt(GM/R) (orbital velocity)

==> w=v/R=sqrt(GM/R)/R = sqrt(GM/R^3) (orbital angular velocity)

3) we know that

w=2*pi/T

==> 2*pi/T= sqrt(GM/R^3)

squaring both sides

4*pi^2/T^2=GM/R^3

==> T^2 = (4*pi^2/GM)* R^3

==> Square of time period is propotional to cube of radius