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A satellite orbiting the moon very near the surface has a period of 110 min. use this information, together with the radius of the moon r=1.74×106m, to calculate the free-fall acceleration on the moon's surface.

Answer :

W0lf93
1.58 m/s^2 For this problem, going to assume that the satellite is VERY close to the surface of a rather unusually smooth moon. In fact, I'm going to assume the satellite is almost skimming the surface, mere millimeters away from it. The relationship between acceleration and velocity for a body in orbit is a = v^2/r Now we've been given r, and we still need v. Since velocity is defined as distance over time, we need to calculate the distance which will be 2 pi r. So 2 * 3.14159 * 1.74x10^6 m = 10932742 m Therefore v will be 10932742 / (110 * 60) = 10932742/ 6600 = 1656 m/s Substituting into the equation for a, gives us a = (1656 m/s)^2/1.74x10^6 m a = 2742336 m^2/s^2 / 1.74x10^6 m a = 1.576055 m/s^2 Rounding to 3 significant figures gives 1.58 m/s^2 Note: If you wish a fun way of learning about orbital mechanics, you could do far worse than playing Kerbal Space Program. As the creator of XKCD, Randall Munroe, says. https://xkcd.com/1356/
johanrusli

The free-fall acceleration on the moon's surface is about 1.58 m/s²

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Further explanation

Newton's gravitational law states that the force of attraction between two objects can be formulated as follows:

[tex]\large {\boxed {F = G \frac{m_1 ~ m_2}{R^2}} }[/tex]

F = Gravitational Force ( Newton )

G = Gravitational Constant ( 6.67 × 10⁻¹¹ Nm² / kg² )

m = Object's Mass ( kg )

R = Distance Between Objects ( m )

Let us now tackle the problem !

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Given:

Period of Satellite = T = 110 min = 6600 s

Radius of Moon = R = 1.74 × 10⁶ m

Asked:

free-fall acceleration on the moon's surface = g = ?

Solution:

[tex]\Sigma F = ma[/tex]

[tex]G \frac{ M m} { R^2 } = m \omega^2 R[/tex]

[tex]G \frac { M } { R^2 } = \omega^2 R[/tex]

[tex]g = \omega^2 R[/tex]

[tex]g = (\frac{2 \pi }{ T} )^2 R[/tex]

[tex]g = ( \frac {2\pi} { 6600 } )^2 \times ( 1.74 \times 10^6 )[/tex]

[tex]g \approx 1.58 \texttt{ m/s}^2[/tex]

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Learn more

  • Impacts of Gravity : https://brainly.com/question/5330244
  • Effect of Earth’s Gravity on Objects : https://brainly.com/question/8844454
  • The Acceleration Due To Gravity : https://brainly.com/question/4189441

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Answer details

Grade: High School

Subject: Physics

Chapter: Gravitational Fields

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