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 0.785 of the escape speed from Earth and (b) its initial kinetic energy is 0.785 of the kinetic energy required to escape Earth

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

f=ma newton's second law 7x8=56n