Question

you will For this problem, you will need to look up physical parameters for objects in space want to keep about 4 significant figures for each. You have a spaceship that is located between the Earth and the Moon. At the spaceship's current location (point A), the net gravitational force on the ship is equal to 0. Determine: a. The distance of the spaceship from the Moon at point A. b. The net gravitational force on the spaceship at point B (the point where the spaceship is twice as far from the Moon as at point A). c. The distance of the spaceship from the Moon at point C (the other point in space where the net gravitational force on the spaceship is the same as at point B). For your physical representation (part 1b), you should graph and label a potential energy diagram showing points A, B, and C.

Answer 1

a) r₁ = 3.836 10⁷ m, b) F = - 3,390 10⁸ N , c) R = 120.3 m

Step-by-step explanation:

a) This is a problem of equilibrium where the force is gravitational, we call F1 the force of the Moon and F2 the force from the earth.

F₁ -F₂ = 0

F₁ = F₂

G m
`M_(m)` / r₁² = G m
`M_(e)` / r₂²

Let's look for the measured distance from a Coordinate System located on the Moon,

r₂ = D - r₁

Where D is the average distance from Terra to the Moon 3.84 10⁸ m

`M_(m)` / r₁² =
`M_(e)` / (D - r₁)²

(D² - 2 D r₁ + r₁²) =
`M_(e)` /
`M_(m)` r₁²

(1 -
`M_(e)` /
`M_(m)`)r₁² - 2D r₁ + D² = 0

Let's replace and solve the second degree equation

(1 - 5.98 10²⁴ / 7.36 10²²) r₁² - 2 3.84 10⁸ r₁ + (3.84 10⁸)² = 0

-80.25 r₁² - 7.68 10⁸ r₁ + 14.75 10¹⁶ = 0

r₁² + 9.57 10⁶ r₁ - 1.838 10¹⁵ = 0

r₁ = [-9.57 10⁶ ±√ (91.58 10¹² + 7.352 10¹⁵)] / 2

r₁ = [-9.57 10⁶ + - 86.28 10⁶] / 2

The results are:

r₁’= 38.355 10 6 m

r₁ ’’ = -47.915 10 6 m

We take the positive result that a distance between the moon and the Earth, the equalization point is r₁ = 3.836 10⁷ m

b) The force at point R = 2 r₁

R = 2 3.8355 10⁷ = 7.671 10⁷ m

F = F₁ - F₂

F = G m
`M_(m)` / R² - G m
`M_(e)` / (D- R)²

F = G m (
`M_(m)` / R² -
`M_(e)` / (D-R)²)

F = m 6.67 10⁻¹¹ (7.36 10²² / (7.671 10⁷)² - 5.98 10²⁴ / (3.84 10⁸ - 0.7671 10⁸)²

F = m 6.67 10⁻¹¹ (0.125076 10⁸ - 0.63329 10⁸)

F = m (-3.3897 10⁸) N

The mass m of the rocket must be known, suppose it is worth 1 kg (m = 1 kg)

F = - 3,390 10⁸ N

c) let's use gravitational force from the moon

F = G m
`M_(m)` / R²

R =√ G m
`M_(m)` / F

R = √ (6.67 10⁻¹¹ 1 7.36 10²² / 3.3897 10⁸)

R = √ (1.4482 10⁴)

R = 1.2034 10² m

R = 120.3 m