Question

If a projectile is launched vertically from the Earth with a speed equal to the escape speed, how high above the earth's surface is it when its speed is half the escape velocity
.5mv^2 -(GMm)/r= .5m(v/2)^2-(GMm)/r

Answer 1

The height of the projectile from the Earth’s surface when its speed is half the escape velocity is $\boxed{19200\,{\text{km}}}$ or $\boxed{1.92 \times {{10}^7}\,{\text{m}}}$.

Further Explanation:

As the projectile is thrown vertically from the surface of the Earth with the escape velocity, the energy of the Earth-projectile system remains conserved at any instant.

The sum of the kinetic energy and the gravitational potential energy of the system remain equal at the surface of Earth as well as at the height where its speed becomes equal to half of escape velocity.

The kinetic energy of the projectile is given as:

$KE = \dfrac{1}{2}m{v^2}$

The gravitational potential energy of the projectile is given as:

$PE = -\dfrac{{GMm}}{r}$

Here, $M$ is the mass of Earth, $r$ is the distance of the particle from the center of Earth.

The escape velocity of Earth is represented as:

${v_e} = \sqrt {\dfrac{{2GM}}{R}}$

Here, ${v_e}$ is the escape velocity and $R$ is the radius of Earth.

The initial energy of the projectile is:

$\begin{aligned}{E_i} &= K{E_i} + P{E_i}\\&= \frac{1}{2}mv_e^2 - \frac{{GMm}}{R} \\\end{aligned}$

The final energy of the projectile is:

$\begin{aligned}{E_f} &= K{E_f} + P{E_f}\\&= \frac{1}{2}m{\left( {\frac{{{v_e}}}{2}}\right)^2} - \frac{{GMm}}{{R + h}}\\\end{aligned}$

Using conservation of energy:

$\begin{aligned}{E_i} &= {E_f} \hfill\\\frac{1}{2}mv_e^2 - \frac{{GMm}}{R} &= \frac{1}{2}m{\left( {\frac{{{v_e}}}{2}} \right)^2} - \frac{{GMm}}{{R + h}} \hfill\\\frac{1}{2}\left( {v_e^2 - \frac{{v_e^2}}{4}} \right) &= GM\left( {\frac{1}{R} - \frac{1}{{R + h}}} \right)\hfill \\\end{aligned}$

Substitute $\sqrt {\dfrac{{2GM}}{R}}$ for ${v_e}$ and simplify the expression for $h$.

$\begin{aligned}\frac{h}{{R + h}}&=\frac{3}{4}\\h&= 3R \\\end{aligned}$

The radius of the Earth is $6400\,{\text{km}}$ . So, the height of the projectile from the surface of Earth is:

$\begin{aligned}h &= 3 \times 6400\,{\text{km}} \\&= 1{\text{9200}}\,{\text{km}}\\&= 1.{\text{92}}{\kern 1pt}\times 1{{\text{0}}^7}\,{\text{m}}\\\end{aligned}$

Thus, the height of the projectile from the Earth’s surface when its speed is half the escape velocity is $\boxed{19200\,{\text{km}}}$ or $\boxed{1.92 \times {{10}^7}\,{\text{m}}}$

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Answer Details:

Grade: Senior School

Subject: Physics

Chapter: Gravitation

Keywords:  Gravitational potential energy, escape speed, half the escape velocity, launched vertically, height of the projectile, conservation of energy.

Answer 2

Answer: h = 3R

Explanation:

Using the law of conservation of energy,

Total energy at the beginning of the launch would be equal to total energy at any point.

kinetic energy + gravitational potential energy = constant

Initial energy of the projectile =$\frac{1}{2}mv_e^2-\frac{GMm}{R}$... (1)

where R is the radius of the Earth, M is the mass ofthe Earth, m is the mass of the projectile.

escape velocity, $v_e=\sqrt{\frac{2GM}{R}}$

Total energy at height h above the Earth where speed of the projectile is half the escape velocity:

$\frac{1}{2}m(v_e/2)^2-G\frac{Mm}{R+h}$ ...(2)

(1)=(2)

⇒$\frac{1}{2}m(v_e/2)^2-G\frac{Mm}{R+h} = \frac{1}{2}mv_e^2-\frac{GMm}{R}$

⇒$G\frac{Mm}{R}-G\frac{Mm}{R+h}= \frac{1}{2}m \frac{3}{4}v_e^2 =\frac{1}{2}m \frac{3}{4} (\sqrt{\frac{2GM}{R}})^2$

⇒$\frac{GM(R+h-R)}{R(R+h)} = \frac{3}{4}(\frac{GM}{R})$

⇒$\frac{h}{R+h} = \frac{3}{4}$

⇒h = 3R

Thus, at height equal to thrice radius of Earth, the speed of the projectile would reduce to half of escape velocity.