Physics, 23.10.2020 18:20 uglybruja81

The acceleration due to gravity at Earth’s surface is 9.8 m/s2. Approximately how far from the surface of Earth would an object need to be to experience half of that acceleration? The radius of Earth is 6,378,100 m, and the mass of Earth is 5.971024 kg. Two identical particles have the same mass, and each holds a charge of 1,60210-19C. What would the mass of each particle need to be for the ratio of the gravitational force to the electrostatic force of the particles to be 1?

Two identical charged objects exert an electrostatic force of 50 N onto each other when they are 125 km apart. What are the charges of the objects?

Two identical objects exert a gravitational force of -0.016 N onto each other when they are 0.25 cm apart. What is the mass of each object?

Answers: 1

Show answers

Answers

Answer from: steve12335

1) 2,641,895.5 m

2) 1.85903476 × 10⁻⁹ kg

3) 9.32 C

4) 38.7 kg

Explanation:

1) The gravitational force of attraction between the object and the mass of the Earth is given by the following formula;

$F} =G\dfrac{M_{1} \cdot m_{2}}{R^{2}} = m_2 \times g$

Where;

M₁ = The mass of the Earth = 5.97 × 10²⁴

m₂ = The mass of the object

G = The universal gravitational constant

R = The radius of the Earth = 6,378,100 m

g = The acceleration due to gravity = 9.8 m/s²

By comparison, we have;

$\dfrac{G \cdot M_{1}}{R^{2}} = g$

When the object is r meters away, g is halved, therefore, we have;

$\dfrac{G \cdot M_{1}}{r^{2}} = \dfrac{g}{2} =\dfrac{G \cdot M_{1}}{\left (2\times R^{2}\right )}$

Therefore, r² = 2·R² = 2 × (6,378,100 m)² = 8.13603192 × 10¹³ m²

r = √(8.13603192 × 10¹³ m²) = 9019995.52217 m

The distance of the object from the surface of the Earth = 9,019,995.52217 - 6,378,100 = 2,641,895.52217 m ≈ 2,641,895.5 m

The object needs to be approximately 2,641,895.5 m from the Earth's surface to experience half of the acceleration due to gravity experienced on the Earth's surface

2) The electrostatic force between the charges = k×q²/r²

Where;

q = 1.602 × 10⁻¹⁹ C

r = Distance between the charges

k = Coulomb constant = 8.9875517923 × 10⁹ kg·m³·s⁻²·C⁻²

We have;

The electrostatic force between the charges = 8.9875517923 × 10⁹ × (1.602 × 10⁻¹⁹)²/r²

The gravitational force between the charges = 6.67408 × 10⁻¹¹ × m²/r²

Given that the ratio between the two forces = 1, we have;

(8.9875517923 × 10⁹ × (1.602 × 10⁻¹⁹)²/r²)/(6.67408 × 10⁻¹¹ × m²/r²) = 1

∴ m² = (8.9875517923 × 10⁹ × (1.602 × 10⁻¹⁹)²)/(6.67408 × 10⁻¹¹) = 3.4560102×10⁻¹⁸ kg²

The mass m = √(3.4560102×10⁻¹⁸ kg²) = 1.85903476 × 10⁻⁹ kg

The mass of each particle will be 1.85903476 × 10⁻⁹ kg

3) F = k×q²/r²

q² = F × r²/k

Which gives;

q² = 50 × 125000²/(8.9875517923 × 10⁹) ≈ 86.93 C²

q = √86.93 ≈ 9.32 C

The charges of each object = 9.32 C

4) $F} =G\dfrac{m^2 }{r^{2}}$

r = 0.25 cm = 0.0025 m

F = 0.016 N

G = 6.67408 × 10⁻¹¹ m³·kg⁻¹·s⁻²

Therefore;

m² = F·r²/G = 0.016 × (0.0025)²/(6.67408 × 10⁻¹¹) ≈ 1498.334 kg²

m = √(1498.334 kg²) ≈ 38.7 kg

The mass of each object = 38.7 kg.

Answer from: kenziepickup

The Coulomb and gravitational laws together with Newton's second law allow to find the results for the questions are:

1) The distance where the acceleration of gravity measures half its value on earth is; r’= 2.64 106 m

2) The mass of the particle is: m = 1.86 10-9 kg

3) The charge of the particles is: q = 9.32 C

4) The mass is: m = 38.7 kg

Part 1

The Universal Gravitational Law says that the force between two objects is proportional to their masses and inversely proportional to the square of their distance

F = $-G \frac{Mm}{r^2}$

Where F is the gravitational force, G the gravitational constant, M and m the mass of the two bodies and r the distance between them.

They ask to calculate the distance to the point where the acceleration is half that of the gravitation acceleration at the earth's surface.

a = g / 2

If we use Newton's second law that gives the relationship between the net force, the mass and the acceleration of the body.

F = m a

Let's substitute

$-G \frac{Mm}{r^2} = m \frac{g}{2}$

r² = $\frac{2GM}{g}$

Let's calculate

r² = $\frac{2 \ 6.67 \ 10^{-11} \ 5.97 \ 10^{24} }{9.8}$

r = $\sqrt{81.360 \ 10^{12}}$

r = 9.02 10⁶ m

This is the distance from the center of the earth, the distance measured from the surface of the planet is

r = R + r '

r ’= r- R

r ’= 9.02 10⁶ - 6.378 10⁶

r’= 2.64 10⁶ m

Part 2

Coulomb's law states that the electric force between two charged bodies is:

F = $k \frac{q_1q_2}{r^2}$

Where q₁ and q₂ are the charges of the leather and k is the Coulomb constant

They ask what mass the particles must have so that the electrostatic force (Fe) and the gravitational force (Fg) have been equal

Indicate the charges of the particles q₁ = q₂ = 1.602 10⁻¹⁹ C

Let's equalize the two forces

$F_e = F_g \\k \frac{q_1q_2}{r^2} = G \frac{m_1m_2}{ r^2 } \\k q^2 = G m^2$

$m = \sqrt{\frac{k}{G} } \ q$

m = $\sqrt{\frac{8.99 \ 10^9 }{ 36.67 \ 10^{-11}} } \ 1.602 \ 10^{-19}$

m = $\sqrt{1.3478 \ 10^{20} } \ 1.602 \ 10^{-19}$

m = 1.86 10⁻⁹ kg

Part 3

They indicate that the electrostatic force between two objects is F = 50 N and the distance between them is r = 125 km = 125 103 m, they ask what is the charge of the objects.

Let's use Coulomb's law

F = $k \frac{q_1q_2}{r^2}$

q² = $\frac{F}{k} \ r^2$

q² = $\frac{50 \ (125 \ 10^3)^2 }{8.99 \ 10^9 }$

q = $\sqrt{86.93}$

q = 9.32 C

Part 4

Indicate the gravitational force between two objects is F_g = -0.016 N when they are separateda distance r = 0.25 cm = 0.25 10⁺² m, ask the mass of the bodies.

Let's use the law of universal gravitation.

$-G \frac{Mm}{r^2} \\m^2 = - \frac{F}{G} \ r^2$