Question

A satellite is projected into space from a point 'P' at a distance '$r_0$' from the centre of the earth with an initial velocity $v_0 = \sqrt{\frac{GM}{r_0}}$ at an angle $\alpha = 37^\circ$ with the horizontal at point 'P' as shown in the figure. It is found that the satellite is propelled into an elliptical orbit. Length of semi-major axis of the ellipse is found to be $\frac{ar_0}{2}$ and eccentricity of the ellipse is e.

Answer 1

e = sin 37° ≈ 0.6

Answer 2

1. Energy:
   At point P,

   $$E = \frac{1}{2}v_0^2 - \frac{GM}{r_0} = \frac{1}{2}\frac{GM}{r_0} - \frac{GM}{r_0} = -\frac{GM}{2r_0}.$$

   Since for an elliptical orbit,

   $$E = -\frac{GM}{2a},$$

   we have $a = r_0$. (This is consistent with the given semi‐major axis $\frac{ar_0}{2}$ if we take the constant $a=2$.)

2. Angular Momentum:
   The horizontal component of the velocity is

   $$v_{0x} = v_0\cos 37^\circ.$$

   Thus,

   $$L = r_0\, v_0 \cos 37^\circ = r_0\sqrt{\frac{GM}{r_0}}\, \cos 37^\circ = \sqrt{GM\,r_0}\, \cos 37^\circ.$$

3. Eccentricity Calculation:
   Using the formula

   $$e = \sqrt{1+\frac{2EL^2}{(GM)^2}},$$

   substitute the values:

   $$2E = -\frac{GM}{r_0}, \quad L^2 = GM\,r_0\,\cos^2 37^\circ.$$

   So,

   $$e = \sqrt{1 + \frac{-\frac{GM}{r_0}\cdot (GM\,r_0\,\cos^2 37^\circ)}{(GM)^2}} = \sqrt{1-\cos^2 37^\circ} = \sin 37^\circ.$$