Question

A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE gives the radial distance (from the Earth's center) the projectile reaches if (a) its initial speed is 0.462 of the escape speed from Earth and (b) its initial kinetic energy is 0.462 of the kinetic energy required to escape Earth

Answer 1

Answer:

(a) h = 1.27 Re

(b) h = 1.86 Re

Explanation:

Let M is the mass of earth and Re is the radius of earth.

initial velocity of projection, v = 0.462 ve

where, ve is the escape velocity of an object on earth surface.

(a)

The value of escape velocity is

$v_{e}=\sqrt{\frac{2GM}{R_{e}}}$

So, $v=0.462\times \sqrt{\frac{2GM}{R_{e}}}$    .... (1)

By using conservation of energy

(Kinetic energy + potential energy ) at the surface of earth = Potential energy at the height h.

where, h is the maximum height upto which the projectile reach

K.E at surface  P.E at surface = P.E at the top

$\frac{1}{2}mv^{2}-\frac{GMm}{R_{e}}=-\frac{GMm}{h}$

By equation (1), substituting the value of v

$\frac{1}{2}\times 0.462^{2}\times \frac{2GM}{R_{e}}-\frac{GM}{R_{e}}=-\frac{GM}{h}$

$\frac{1}{2}\times 0.462^{2}\times \frac{2}{R_{e}}-\frac{1}{R_{e}}=-\frac{1}{h}$

h = 1.27 Re

(b)

initial kinetic energy = 0.462 times the kinetic energy required to escape

$\frac{1}{2}mv^{2}=0.462\times \frac{1}{2}mv_{e}^{2}$

$\frac{1}{2}mv^{2}=0.462\times \frac{1}{2}m\times \frac{2GM}{R_{e}}$

So, again by using the conservation of energy

Kinetic energy at the surface + Potential energy at the surface = Potential energy at the top

$0.462\times \frac{1}{2}m\times \frac{2GM}{R_{e}}-\frac{GMm}{R_{e}}=-\frac{GMm}{h}$

$0.462\times \frac{1}{R_{e}}-\frac{1}{R_{e}}=-\frac{1}{h}$

h = 1.86 Re