Question

Consider a satellite of mass ms in a circular earth orbit. Anysatellite orbiting in the upper reaches of our atmosphere willexperience a small drag force, F_d, opposite to itsdirection of motion. This drag force will cause the satellite tospiral inward toward smaller orbital radii. In the process, theorbital velocity of the satellite actually increases! This problemwill investigate how a drag force can cause a body in orbit tospeed up.

The situation is sketched in Fig. 1. The diagram shows separateforce and velocity diagrams for the satellite. The mass of theearth is denoted me, while the mass of the satelliteis ms. The tangential velocity of the satellite isv_t and the small radial velocity is acquires because itis spiraling inwards is v_r. The size of v_r isgreatly exaggerated in the diagram, and we can take v_r<< v_t and use v_t as an adequateapproximation to the magnitude of the total velocity, v. We willcall θ the ‘angle of de- scent’.

Turning to the force diagram, we see that the total force actingon the satellite is the vector sum of F⃗_d andF⃗_g. Again, the drag force is much weaker than thegravitational force, so F_d << F_g . Wecan use this to our advantage by just considering the effect ofF_g along the line of descent. We can see from thediagram that F_g has a component along the line ofdescent given by F_g sin(θ).

(a) First show how we arrive at the following two basicequations governing this system, and carefully describe thephysical meaning of the quantity m_sa_d inwords:

F_gsinθ−F_d = m_sa_d (1)

F_g = (m_s*v_t²)/r= G(m_sm_e)/r² (2)

(b) We can find out more about a_d and its relation tothe various forces and velocities by powerful

use of the chain rule for differentiation:

Explain Eq. 3 in your own words. Where did vr comefrom and why is the ≃ sign there?

a_d = dv/dt = (dv/dr)*(dr/dt) =v_r(dv/dr) ≃ v_r(dv_t/dr) (3)

(c) We can now make use of Eqs. 2 and 3 to show that:

m_sa_d = (1/2)F_g sinθ