Question

(Superposition, quadratic formuła) The Earth (mass M) is at origin, and the Moon (mass m ) is located at a distance d. (a) Using Newton's law of gravitation F, find the total gravitational force acting on a spaceship (mass m) at r. (b) Find the location(s) where this force vanishes, assuming that d -380,000 km. 17. (Differentiation, integration) What distance do you cover in the x-direction per unit time if your speis given by: (a) ?

Answer 1

Answer:

a)    Fa = G m2 [M / r² - m / (d-r)²]  

b) r2 = 31 10⁶ m

Explanation:

The equation of the law of universal gravitation is

            F = G m1 m1/ r²

The value of the gravitation constant is 6.67 10-11 N m²/kg². This force is always attractive.

Let's calculate the value of that force on the spacecraft, add the strength

Earth's force to the ship

            F1 = G M m2 / r²

The moon force ship

            F2 = G m m2 / (d-r)²

Total force is

            Fa = F1 — F2

            Fa = G M M2 / r² - G m m2 / (d-r)²

            Fa = G m2 [M / r² - m / (d-r)²]

This is the force on the spaceship

b)  Let's look for the point where the force is zero, for this we can see that the value of the bracket must be zero

           Fa = 0

           [M / r² - m / (d-r)²] = 0

            M / r² = m / (d-r)²

           (d-r)² = m/M   r²

           d² -2rd + r² - m/M   r² = 0

           r² [m/M - 1] + r 2d - d² = 0

This is a second degree equation for r, we solve the to find the results.

           r = {-2d ±√[4d² - 4 [m/M -1] (-d²)]} / (2 [m/M-1])

           r = {-2d ± √ [4d² (1 + (m/M-1)]} / 2(m/M-1)

           r = {-2d ± 2d √(m/M)}  / (2(m/M-1))

           r = 2d {-1 ± √(m/M)} / 2(m/M-1)

           r = d [-1 ± √(m/M)] /  (m/M-1)

To find the explicit value we substitute the values ​​that we can find in tables

          m = 7.36 1022 kg

          M = 5.98 1024 kg

          d = 380000 km (1000m / 1 km) = 380 10⁶ m

          r = 380 10⁶ [-1 ±√(7.36 10²² / 5.98 10²⁴)] /(7.36 10²² / 5.98 / 10²⁴ -1)

          r = 380 10⁶ [-1 - √ (1.23 10²)] / (123-1)

          r = 380 106 [-1 ± 11] / 122

 

          r1 = 380 10⁶ 10/122 = 380 10⁶ 0.08197

          r1 =  31 10⁶ m

          r2 = 380 106 [-12/122] = 380 10 6 0.09836

          r2 = -37 10⁶ m

The correct distance is the positive r2 = 31 10⁶ m

c) let's use Newton's second law, to find the acceleration in the spacecraft

 

          F = m a

          a = Fa / m2 = G m2 [M / r² - m / (d-r)²] / m2

          a = G [M/r² - m/(d-r)²]

Since we have acceleration, we can use the definition of kinematics

           a = dv / dt = dv / dr dr / dt = dv / dr v

           v dv = a dr

           v dV = G [M /r² - m /(d-r)²] dr

We integrate

            ½ (V² - Vo²) = G [M (-1 /r) -m (1 / (d-r)

We evaluate between the initial point where we can assume that the initial velocity is zero for the Xo position and the final point with velocity v at the points

            V² = 2G [M (1 / Xo - 1 /X) - m (1 / (d-X) - 1 /(d-xo)]

            V² = [-M /X -m /(d-X)] 2G + constant

We now use the definition of speed

            v = dx / dt

            dx = V dt

We substitute, perform the integral and simplify, if we can make the constant zero

             dx = √([-M / X -m / (d-X)] 2G) dt

             dx / √([-M / X -m / (d-X)] 2G = dt