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. What is the moon's acceleration due to gravity?

Respuesta :

To solve this problem it is necessary to apply the concepts related to the centripetal Force, the Force of the weight produced by the gravity of the star and the definition of the velocity as a function of time (period)

In other words, for balance to exist, the force acting on the body (weight) must be equal to the centripetal force that attracts it in motion. That is to say

[tex]\sum F = 0[/tex]

[tex]mg_{moon}-ma_r=0[/tex]

[tex]mg_{moon}=ma_r[/tex]

[tex]g_{moon} = \frac{v^2_{sat}}{r}[/tex]

The value of the given period is 110min or 6600s.

At the same time we know that the speed of a body depending on its period (simple harmonic movement) is subject to

[tex]v_{sat} = \frac{2\pi r}{T}[/tex]

[tex]v_{sat} = \frac{2\pi (1.74*10^6)}{6600s}[/tex]

[tex]v_{sat} = 1.66*10^3m/s[/tex]

Finally, replacing this value in the event of gravity we have to

[tex]g_{moon} = \frac{v^2_{sat}}{r}[/tex]

[tex]g_{moon} = \frac{(1.66*10^3)^2}{1.74*10^6}[/tex]

[tex]g_{moon} = 1.58m/s^2[/tex]

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