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 0.241 of the escape speed from Earth and (b) its initial kinetic energy is 0.241 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?

Answer 1

Answer:

a. 1.06R b. 1.32R c. GMm/R

Explanation:

a. Considering the conservation of mechanical energy,

K₁ + U₁ = K₂ + U₂     (1)where K₁ and K₂ = initial and final kinetic energies of projectile, and U₁ and U₂ = initial and final potential energies of projectile

K₁ = 1/2mv²,   K₂ = 0,  U₁ = -GMm/R   where R = radius of earth , U₂ = '-GMm/r where r = radius at maximum height.

So, inputting the variables into (1), we have

1/2mv² - GMm/R = 0 - GMm/r

1/2mv² = -GMm/r + GMm/R  (2)

Now v = 0.241v₁    where v₁ =√(2GM/R) escape velocity

Substituting v into (2) above, we have

1/2m[0.241√(2GM/R)]² = -GMm/r + GMm/R

0.058GMm/R = -GMm/r + GMm/R

GMm/r = GMm/R -'0.058GMm/R

GMm/r = 0.942GMm/R

r = R/0.942

r = 1.06R

b. When K₁ = 0.241K wh6ere K = escape kinetic energy = GMm/R. So

K₁ + U₁ = K₂ + U₂

0.241GMm/R -'GMm/R = 0 -' GMm/r

-'0.759GMm/R = '-GMm/r

r = R/0.759 = 1.32R

c. If it is to escape earth, its initial velocity must equal the escape velocity.

So its least initial mechanical energy is its escape kinetic energy

1/2m[√(2GM/R)]² = GMm/R

Answer 2

Answer:

(a) r = 1.062·R$_E$ = $\frac{531}{500} R_E$

(b) r = $\frac{33}{25} R_E$

(c) Zero

Explanation:

Here we have escape velocity v$_e$ given by

$v_e =\sqrt{\frac{2GM}{R_E} }$ and the maximum height given by

$\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r}$

Therefore, when the initial speed is 0.241v$_e$ we have

v = $0.241\times \sqrt{\frac{2GM}{R_E} }$ so that;

v² = $0.058081\times {\frac{2GM}{R_E} }$

v² = ${\frac{0.116162\times GM}{R_E} }$

$\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r}$ is then

$\frac{1}{2} {\frac{0.116162\times GM}{R_E} }-\frac{GM}{R_E} = -\frac{GM}{r}$

Which gives

$-\frac{0.941919}{R_E} = -\frac{1}{r}$ or

r = 1.062·R$_E$

(b) Here we have

$K_i = 0.241\times \frac{1}{2} \times m \times v_e^2 = 0.241\times \frac{1}{2} \times m \times \frac{2GM}{R_E} = \frac{0.241mGM}{R_E}$

Therefore we put  $\frac{0.241GM}{R_E}$ in the maximum height equation to get

$\frac{0.241}{R_E} -\frac{1}{R_E} =-\frac{1}{r}$

From which we get

r = 1.32·R$_E$

(c) The we have the least initial mechanical energy, ME given by

ME = KE - PE

Where the KE = PE required to leave the earth we have

ME = KE - KE = 0

The least initial mechanical energy to leave the earth is zero.