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Determine the orbits period of the moon when the distance between the earth and the moon is 3.82 x 10 to the power of 8

Sagot :

spectrojamz

Answer

From

V=Distance/time

The distance round a circular path or Object is 2πr

so

v=2πr/T

Making T(period) subject

T=2πr/v

where v is the linear velocity.

We don't have the velocity but we can get it.

The Moon and Earth exert gravitational force on each other and the Moon is kept in Orbit by Centripetal force.

Since the Moon doesn't fly out of Orbit... it must mean that the Gravitational force being exerted on it by the Earth is equal to its centripetal Force.

Equating Both (Fg=Fc)

Fg=GMm/R²

Fc=mv²/r

GMm/R² = mv²/R

Canceling out "m" and "r" on both sides

we're left with

V²=GM/R

V=√GM/R

Where M= Mass of Earth (5.98x10^24)

R=Distance between the center of earth and the Moon

G= Gravitational Constant(Value 6.67x10^-11)

V=√6.67x10^-11 x 5.98x10^24/(3.82x10^8)

V=1021.84meters per second.

Now

T=2πr/v

=2π x 3.82x10^8 / 1021.84

T=2.35x10^6seconds

Converting to days by dividing by(24 x 3600)

You have

T=27.2days approx.

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