Physics, 31.12.2019 05:31 jess4791

At some point along the direct path from the center of the earth to the center of the moon, the gravitational force of attraction on a spacecraft from the moon becomes greater than the force from the earth. a) how far from the center of the earth does this occur? b) at this location, how far is the spacecraft from the surface of the moon? c) how far is it from the surface of the earth?

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Answer from: erinmcdonaldr6860

a) r₁ = 3.4565 10⁸ m , b) $H_{m}$ = 3.661 10⁷ m , c) $H_{e}$ = 3.39 10⁸ m

Explanation:

a) For this exercise we will use Newton's second law where the forces are of gravitational attraction

F = G m M / r²

Let's locate the point where the forces are equal and from here we know which way they are greater

F₁ - F₂ = 0

F₁ is the force of attraction from Earth and F₂ the force of attraction from the Moon, let's use the subscript "e" for the Earth and the subscript "m" for the Moon

G m $M_{e}$ / r₁² = G m $M_{m}$ / r₂²

$M_{e}$ / $M_{m}$ = (r₁ / r₂)²

Let's measure the distance from the same reference system, which we will place in the center of the Earth

r₁ = r₁

r₂ = D - r₁

Where D is the distance from beats to the Moon, substitute

r₁ / D-r₁ = √ $M_{e}$ / $M_{m}$

r₁ / D-r₁ = √ (5.98 1024 / 7.36 1022) = 9.0139

r₁ = 9.0139 (D -r₁)

r₁ (1+ 9.0139) = 9.0139 D

r₁ = 9.0139 / 10.0139 D

r₁ = 0.9001 3.84 10⁸

r₁ = 3.4565 10⁸ m

It is this point sides forces are equal for greater distances the force of the moon exceeds the attraction of the Earth

b) The radius of the moon is

$R_{m}$ = 1.74 10⁶ m

The distance measured from the Moon is

r₂ = D -r₁

r₂ = 3.84 108 - 3.4565 108

r₂ = 0.3835 108 m

r₂ = 3.835 107 m

The distance r₂ is measured from the center of the Moon, the distance measured from the surface is

$H_{m}$ = r₂ - $R_{m}$

$H_{m}$ = 3.835 10⁷ - 1.74 10⁶

$H_{m}$ = 3.661 10⁷ m

c) The distance from the Earth's surface is

$H_{e}$ = r1 - $R_{e}$

$H_{e}$ = 3.4565 10⁸ - 6.37 10⁶

$H_{e}$ = 3.39 10⁸ m

Answer from: Quest

i already answered it by 34/s

Answer from: Quest

Answer: a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same sign as the -4.0 µc charge.======electricstatic force between two point charges is given by $f = \dfrac{kq_1q_2}{r^2}$ where k is coulomb's constant, 8.99×10⁹ n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ = 6.0 µc, q₂ = -4.0 µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1 µc = 10⁻⁶ c) $f_{01} = \dfrac{kq_0q_1}{(\text{6.0 m})^2}$ the force from q₂ $f_{02} = \dfrac{kq_0q_2}{(\text{15 m})^2}$ the force from q₃ $f_{03} = \dfrac{kq_0q_1}{(\text{26 m})^2}$ the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃ $\begin{aligned} f_{net} & = 0\\ f_{01} + f_{02} + f_{03} & = 0 \\ \dfrac{kq_0q_1}{(\text{6.0 m})^2} + \dfrac{kq_0q_2}{(\text{15 m})^2} + \dfrac{kq_0q_3}{(\text{26 m})^2} & = 0 \\ kq_0 \left( \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2}\right) & = 0 \\ \dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2} + \dfrac{q_3}{(\text{26 m})^2} & = 0 \end{aligned}$ isolating: $\begin{aligned} \dfrac{q_3}{(\text{26 m})^2} & = -\dfrac{q_1}{(\text{6.0 m})^2} - \dfrac{q_2}{(\text{15 m})^2} \\ q_3 & = -(\text{26 m})^2\left(\dfrac{q_1}{(\text{6.0 m})^2} + \dfrac{q_2}{(\text{15 m})^2}\right) \\ q_3 & = -(\text{26 m})^2\left(\dfrac{6.0\times10^{-6} \text{ c}}{(\text{6.0 m})^2} + \dfrac{-4.0\times10^{-6} \text{ c}}{(\text{15 m})^2}\right) \\ q_3 & = -1.0064888 \times 10^{-4}\text{ c} \\ q_3 & = -1.0\times 10^{-4}\text{ c} \end{aligned}$ a charge of -1.0 × 10⁻⁴ c must be placed at x = 26 m.the sign of the charge will be the same as the -4.0 µc charge