As we know, the moon is a satellite of our earth, what is the
theoretical period of the moon? The average radius of the
moon's orbit is 3.84 108 m and the mass of the earth is 5.97 x
1024 kg (in hours, G = 6.67 x 10-9 N (m/kg) 3).

As We Know, The Moon Is A Satellite Of Our Earth, What Is Thetheoretical Period Of The Moon? The Average

Answers

Answer 1

Answer:c

Explanation:c

Answer 2

This question involves the concepts of the time period, orbital radius, and gravitational constant.

The theoretical period of the moon is "658 hr".

The theoretical time period of the moon around the earth can be found using the following formula:

[tex]\frac{T^2}{R^3}=\frac{4\pi^2}{GM}[/tex]

where,

T = Time Period of Moon = ?

R = Orbital Radius = 3.84 x 10⁸ m

G = Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of Earth = 5.97 x 10²⁴ kg

Therefore,

[tex]\frac{T^2}{(3.84\ x\ 10^8\ m)^3}=\frac{4\pi^2}{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(5.97\ x\ 10^{24}\ kg)}\\\\T^2=(9.91\ x\ 10^{-14}\ s^2/m^3)(56.62\ x\ 10^{24}\ m^3)\\\\T=\sqrt{561.34\ x\ 10^{10}\ s^2}[/tex]

T = 2.37 x 10⁶ s[tex](\frac{1\ h}{3600\ s})[/tex]

T = 658 hr

Learn more about the orbital time period here:

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The attached picture shows the derivation of the formula for orbital speed.

As We Know, The Moon Is A Satellite Of Our Earth, What Is Thetheoretical Period Of The Moon? The Average

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Answer:

4620

Explanation:

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What is the Solar System?

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Answer:

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Answers

Answer:

The answer is "[tex]5.184 \times 10^{23}[/tex] kg"

Explanation:

Given:

[tex]g= 6.003 \ \frac{m}{s^2}\\\\G= 6.67 \times 10^{-11} \ \frac{N - m^2}{kg^2}\\\\M=?\\\\radius= 2,400 \ km\\\\[/tex]

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[tex]Diameter =radius \times 2[/tex]

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[tex]\to g = G \times \frac{M}{(\frac{d^2}{4})}\\\\\to g=G\times \frac{4M}{d^2}\\\\\to 4M= \frac{g \times d^2}{G} \\\\\to M= \frac{g \times d^2}{4G} \\\\[/tex]

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Answer:

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Answer:

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Answer:

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Explanation:

Given that a stone dropped from the top of a 80m high building strikes the ground at 40 m/s after falling for 4 seconds. The stone's potential energy with respect to the ground is equal to its kinetic energy. (use 9 - 10 m/s)

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Answer:

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Answers

Answer:

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Answer:

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Explanation:

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Answer

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Answer:

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A :-) KE = 1 by 2 mv^2
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