Answer:
Explanation:
Given that,
We have a projectile that is shot from the earth
Then, it's escape velocity is
Ve = √(2GM / R)
Ve² = 2GM / R
M is the mass of earth
R is the radius of Earth
Using conservation of energy
Ki + Ui = Kf + Uf
The final kinetic energy is zero since the body is finally at rest
Then,
Ki + Ui = Uf
Kinetic energy can be determine using
K = ½mv²
Potential energy can be determine using
U = -GmM / R
Where m is mass of the body
Then,
Ki + Ui = Uf
½mv² - GmM / R = -GmM / r
r is the maximum height reached
m cancel out
½ v² - GM / R = -GmM / r
A. We want to find the maximum height reached when initial speed (v) is equal to 0.785 escape velocity
So, v = 0.785•Ve
So,
½v² - GM / R = -GM / r
½(0.785•Ve)² - GM / R = -GM / r
0.308•Ve² - GM / R = -GM / r
0.308 × 2GM / R - GM / R = -GM / r
Divide through by GM
0.616 / R - 1 / R = -1 / r
-0.384 / R = -1 / r
Cross multiply
-0.384r = -R
r = -R / -0.384
r = 2.6 R
B. When it initial kinetic energy is is 0.785 of the kinetic energy required to escape Earth
Ki = 0.785 Ke
Ki = 0.785 ½mVe²
Ki = 0.785 × ½m× 2GM / R
Ki = 0.785•G•m•M / R
So,
Ki + Ui = Uf
0.785•G•m•M / R - GmM / R = -GmM / r
Let G•m•M cancels out
0.785 / R - 1 / R = -1 / r
-0.215 / R = -1 / r
Cross multiply.
-0.215r = -R
r = -R / -0.215
r = 4.65 R
