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3 years ago

How far does the moon move away from earth every year?

Physics

2 answers:

Mamont248 [21]3 years ago

6 0

Answer:

1.6 inch (4 cm)

Explanation:

Currently moon is about 3.85 ×10⁵ km away from the Earth on an average. The gravitational pull of the Earth is keeping the moon in its orbit around the Earth. The gravitational pull of moon causes tidal bulge. The Earth tends to drag ahead this bulge from the slow orbiting moon. Due to this, the rotation of the Earth is slowing down and moon is being pushed away slowly. Every year moon is pushed to about 1.6 inches or 4 cm away from the Earth.

Oxana [17]3 years ago

4 0

The moon is moving away from Earth at a rate of approximately 3.78 cm per year.
This migration of the Moon from the Earth is mainly due to the action of the Earth tides. It can be explained as follows:
- the Moon exerts a gravitational force on the Earth, which is stronger at the Equator (since the Equator is closer to the Moon), creating the tides
- However, the Earth rotates faster on its axis (one rotation every 24 hours) than the Moon (one rotation every 27 days), therefore the tidal bulge on Earth tries to pull the Moon "ahead" in its orbit. As a result, the Moon tends to sped up.
<span>- As opposite reaction, the Earth tends to slow down in its rotation, with a loss of angular momentum. Since the angular momentum must be conserved, the radius of the orbit of the Moon becomes larger, and this explains why the Moon is moving away from the Earth.</span>

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Nataliya [291]

Answer:

∅ = 89.44°

Explanation:

In situations like this air resistance are usually been neglected thereby making g= 9.81 m/ $s^{2}$

Bring out the given parameters from the question:

Initial Velocity ( $V_{1}$ ) = 1000 m/s

Target distance (d) = 2000 m

Target height (h) = 800 m

Projection angle ∅ = ?

Horizontal distance = $V_{1x}$ tcos ∅ .......................... Equation 1

where $V_{1x}$ = velocity in the X - direction

t = Time taken

Vertical Distance = y = $V_{1y}$ t - $\frac{1}{2}$ g $t^{2}$ ................... Equation 2

Where $V_{1y}$ = Velocity in the Y- direction

t = Time taken

$V_{1y}$ = $V_{1}$ sin∅

Making time (t) subject of the formula in Equation 1

t = d/( $V_{1x}$ cos ∅)

t = $\frac{2000}{1000coso}$ = $\frac{2}{cos0}$ = $\frac{d}{cos o}$ ...................Equation 3

substituting equation 3 into equation 2

Vertical Distance = d = $V_{1y}$ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = sin∅ $\frac{d}{cos o}$ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Vertical Distance = h = dtan∅ - $\frac{1}{2}$ g $\frac{2}{cos0} ^{2}$

Applying geometry

$\frac{1}{cos o}$ = $tan^{2} o$ + 1

Vertical Distance = h = d tan∅ - 2 g ( $tan^{2} o$ + 1)

substituting the given parameters

800 = 2000 tan ∅ - 2 (9.81)( $tan^{2} o$ + 1)

800 = 2000 tan ∅ - 19.6( $tan^{2} o$ + 1) Equation 4

Replacing tan ∅ = Q .....................Equation 5

In order to get a quadratic equation that can be easily solve.

800 = 2000 Q - 19.6 $Q^{2}$ + 19.6

Rearranging 19.6 $Q^{2}$ - 2000 Q + 780.4 = 0

$Q_{1}$ = 101.6291

$Q_{2}$ = 0.411

Inserting the value of Q Into Equation 5

tan ∅ = 101.63 or tan ∅ = 0.4114

Taking the Tan inverse of each value of Q

∅ = 89.44° ∅ = 22.37°

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Answer:

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