A projectile is shot directly away from Earth’s surface.Neglect the rotation of Earth. What multiple of Earth’s radius gives the radial distance a projectile reaches if (a) its initial speed is $\mathbf{0}\mathbf{.}\mathbf{500}$ of the escape speed from Earth and(b) its initial kinetic energy is $\mathbf{0}\mathbf{.}\mathbf{500}$ of the kinetic energy required to escape Earth?(c)What is the least initial mechanical energy required at launch if the projectile is to escape Earth?

The initial speed of the projectile is $0.500$of the escape speeds from Earth.

The initial kinetic energy of the projectile is $0.500$ of the kinetic energy required to escape Earth.

Using the principle of energy conservation,find themultiple of Earth’s radius which gives the radial distance a projectile reaches if its initial speed is of the escape speed from Earth and if its initial kinetic energy is of the kinetic energy required to escape Earth.According to the law ofconservation of energy, energy can neither be created nor be destroyed.

Formulae are as follows:

$$\begin{gathered} K_i+U_i=K_f+U_f \\ U=-\frac{GMm}{R} \\ K=\frac{1}{2}m{v}^2 \end{gathered}$$

where,M, and m are masses, R is the radius, v is velocity, G is gravitational constant, K is kinetic energy and U is potential energy.

Now,

$K_i+U_i=K_f+U_f$

As

$U_i=-\frac{GMm}{R_E}$and$U_f=-\frac{GMm}{R},$

$$\begin{gathered} K_i=\frac{1}{2}m{v}^2\& K_f=0 \\ \frac{1}{2}m{v}^2-\frac{GMm}{R_E}=0-\frac{GMm}{R} \end{gathered}$$

As

$$\begin{gathered} v_e=\sqrt{\frac{2GM}{R_E}} \\ \frac{1}{8}m{\left(\sqrt{\frac{2GM}{R_E}}\right)}^2-\frac{GMm}{R_E}=0-\frac{GMm}{R} \\ \frac{2GMm}{8R_E}-\frac{GMm}{R_E}=-\frac{GMm}{R} \\ \frac{GMm}{4R_E}-\frac{GMm}{R_E}=-\frac{GMm}{R} \\ \frac{3}{4R_E}=\frac{1}{R} \\ \frac{R_E}{R}=\frac{3}{4} \\ \frac{R}{R_E}=\frac{4}{3} \end{gathered}$$

$R=1.33R_E$

Therefore, the multiple of Earth’s radius gives the radial distance a projectile reaches if its initial speed is of the escape speed from Earth is role="math" localid="1661194870557" $1.33R_E$.

Now,

$K_i+U_i=K_f+U_f$

As

.$$\begin{gathered} U_i=-\frac{GMm}{R_E} \\ {U}_f=-\frac{GMm}{R} \\ {K}_i=\frac{1}{2}m{v}^2 \\ {K}_f=0 \\ \frac{1}{2}m{v}^2-\frac{GMm}{R_E}=0-\frac{GMm}{R} \end{gathered}$$

As

$$\begin{gathered} \frac{1}{2}m{v}^2=\frac{\left(\frac{1}{2}m{v}_e^2\right)}{2} \\ \frac{\left(\frac{1}{2}m{v}_e^2\right)}{2}-\frac{GMm}{R_E}=0-\frac{GMm}{R} \\ \frac{m{v}_e^2}{4}-\frac{GMm}{R_E}=0-\frac{GMm}{R} \end{gathered}$$

As

$$\begin{gathered} v_e=\sqrt{\frac{2GM}{R_E}} \\ \frac{1}{4}m{\left(\sqrt{\frac{2GM}{R_E}}\right)}^2-\frac{GMm}{R_E}=0-\frac{GMm}{R} \\ \frac{2GMm}{4R_E}-\frac{GMm}{R_E}=-\frac{GMm}{R} \\ \frac{1}{2}\frac{GMm}{R_E}-\frac{GMm}{R_E}=-\frac{GMm}{R} \\ -\frac{1}{2}\frac{GMm}{R_E}=-\frac{GMm}{R} \\ \frac{R}{R_E}=2 \\ R=2R_E \end{gathered}$$

Therefore, the multiple of Earth’s radius $R_E$ that gives the radial distance a projectile reaches if its initial kinetic energy is $0.500$ of the kinetic energy required to escape Earth.

Now,

$K_i+U_i=K_f+U_f$

As

$$\begin{gathered} U_i=-\frac{GMm}{R_E} \\ {\text{U}}_{\text{f}}=0, \\ {K}_i=\frac{1}{2}m{v}^2 \\ {K}_f=0 \\ \frac{1}{2}m{v}^2-\frac{GMm}{R_E}=0-0 \\ \frac{1}{2}m{v}_e^2-\frac{GMm}{R_E}=0 \end{gathered}$$

As

$$\begin{gathered} v_e=\sqrt{\frac{2GM}{R_E}} \\ \frac{m}{2}{\left(\sqrt{\frac{2GM}{R_E}}\right)}^2-\frac{GMm}{R_E}=0 \\ \frac{GMm}{R_E}-\frac{GMm}{R_E}=0 \\ 0=0 \end{gathered}$$

Hence, the least initial mechanical energy required at launch if the projectile is to escape Earth is 0 .

Therefore, using the formula for gravitational potential energy and kinetic energy along with the law of conservation of energy, the required distance can be found.