What is the net force necessary for a 1.6 x 103 kg car to accelerate forward at 2.0<br> m/s?

PART ONE

stira [4]

Answer:

3.64×10⁸ m

3.34×10⁻³ m/s²

Explanation:

Let's define some variables:

M₁ = mass of the Earth

r₁ = r = distance from the Earth's center

M₂ = mass of the moon

r₂ = d − r = distance from the moon's center

d = distance between the Earth and the moon

When the gravitational fields become equal:

GM₁m / r₁² = GM₂m / r₂²

M₁ / r₁² = M₂ / r₂²

M₁ / r² = M₂ / (d − r)²

M₁ / r² = M₂ / (d² − 2dr + r²)

M₁ (d² − 2dr + r²) = M₂ r²

M₁d² − 2dM₁ r + M₁ r² = M₂ r²

M₁d² − 2dM₁ r + (M₁ − M₂) r² = 0

d² − 2d r + (1 − M₂/M₁) r² = 0

Solving with quadratic formula:

r = [ 2d ± √(4d² − 4 (1 − M₂/M₁) d²) ] / 2 (1 − M₂/M₁)

r = [ 2d ± 2d√(1 − (1 − M₂/M₁)) ] / 2 (1 − M₂/M₁)

r = [ 2d ± 2d√(1 − 1 + M₂/M₁) ] / 2 (1 − M₂/M₁)

r = [ 2d ± 2d√(M₂/M₁) ] / 2 (1 − M₂/M₁)

When we plug in the values, we get:

r = 3.64×10⁸ m

If the moon wasn't there, the acceleration due to Earth's gravity would be:

g = GM / r²

g = (6.672×10⁻¹¹ N m²/kg²) (5.98×10²⁴ kg) / (3.64×10⁸ m)²

g = 3.34×10⁻³ m/s²

4 0

3 years ago

A satellite of mass M = 270kg is in circular orbit around the Earth at an altitude equal to the earth's mean radius (6370 km). A

zubka84 [21]

To solve this problem we will apply the concepts related to Orbital Speed as a function of the universal gravitational constant, the mass of the planet and the orbital distance of the satellite. From finding the velocity it will be possible to calculate the period of the body and finally the gravitational force acting on the satellite.

PART A)

$V_{orbital} = \sqrt{\frac{GM_E}{R}}$