For a particle projected in a transverse direction from a height h above earth's surface, find the minimum initial velocity so that it just grazes the surface of earth such that path of this particle would be an ellipse with centre of earth as the farther focus, point of projection as the apogee and a diametrically opposite point on earth's surface as perigee.

Open in App

Solution

Suppose velocity of projection at point A is $v_{A}$ and at point B, the velocity of the particle is $v_{B}$ .
Then applying Newton's second law at points A and B, we get
$\frac{m v_{A}^{2}}{ρ_{A}} = \frac{G M_{e} m}{(R + h)^{2}}$ and $\frac{m v^{2}}{ρ_{B}} - \frac{G M_{e} m}{R^{2}}$
where $ρ_{A}$ and $ρ_{B}$ are radii of curvature of the orbit at points A and B of the ellipse, but $ρ_{A} = ρ_{B} = ρ (s a y)$ .
Now applying conservation of energy at points A and B,
$\frac{- G M_{e} m}{R + h} + \frac{1}{2} m v_{A}^{2} = - \frac{G M_{e} m}{R} + \frac{1}{2} m v_{B}^{2}$
$\Rightarrow G M_{e} m (\frac{1}{R} - \frac{1}{(R + h)}) = \frac{1}{2} (m v_{B}^{2} - m v_{A}^{2})$
$= (\frac{1}{2} ρ G M_{e} m (\frac{1}{R^{2}} - \frac{1}{(R + h)^{2}}))$
or $ρ = \frac{2 R (R + h)}{2 R + h} = \frac{2 R r}{R + r}$
Thus we get
$V_{A}^{2} = \frac{ρ G M_{e}}{(R + h)^{2}} = 2 G M_{e} \frac{R}{r (r + R)}$
where r $=$ distance of point of projection from earth's centre $= R + h$ .

Suggest Corrections

Similar questions

m

v_{0}

Θ

h

v

v

4 R

V_{1}

30^{\circ}

\frac{G M}{R} = 6.4 \times 10^{7} m^{2} / s^{2}

V_{1}

3 R

R

Join BYJU'S Learning Program