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A moon of mass 1 x 10" kg is in a circular orbit around a planet. The planet exerts a gravitational force of 2 x 104 N on the moon. The centripetal acceleration of the moon is most nearly

Respuesta :

C. 20 m/s²

The centripetal acceleration of the moon is nearly equal to 20 m/s²

Explanation:

Given:

Mass of the moon : [tex]m = {1 \times 10^\(20}\ kg[/tex]

Gravitational force : [tex]F_G = 2 \times 10^\(21 \)N[/tex]

[tex]a_c = ?[/tex]

As the gravitational force provides the centripetal force necessary for the moon to perform circular motion around the planet;

Centripetal Force = Gravitational force

[tex]F_C = F_G[/tex] .....................(1)

[tex]Centripetal force \ F_C= m\times a_c \\ where\\ F_C = Centripetal force \\a_c= Centripetal \ acceleration \\ m=mass\ of \ the \ moon[/tex]

From (1)

[tex]F_C = F_G = m \times a_c\\\\a_c = \frac{F_G}{m} = \frac{2 \times 10^\(21}{1 \times 10^\(20}[/tex]

[tex]a_c = 2 \times 10 = 20 \ m/s^2[/tex]

Centripetal acceleration = 20 m/s²

aksnkj

Answer:

The centripetal acceleration of the moon is [tex]20\; \rm m/s^2[/tex].

Explanation:

Given information:

Mass of the moon is [tex]m=1\times10^{20} \texttt{ kg}[/tex]

The gravitational force exerted by the planet on the moon is, [tex]F=2\times10^{21}\rm \;N[/tex].

Now, the moon is revolving around the planet and the gravitational force will be responsible for centripetal force on the moon.

The centripetal force on the moon will be,

[tex]F_c=ma_c[/tex]

where [tex]a_c[/tex] is the centripetal acceleration of the moon.

So, the centripetal acceleration of the moon will be calculated as,

[tex]F=F_c\\2\times10^{21}=ma_c\\2\times10^{21}=1\times10^{20}\times a_c\\a_c=20\; \rm m/s^2[/tex]

Therefore, the centripetal acceleration of the moon is [tex]20\; \rm m/s^2[/tex].

For more details, refer the link:

https://brainly.com/question/14249440?referrer=searchResults

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