contestada

The free-fall acceleration at the surface of planet 1 is 26 m/s^2 . The radius and the mass of planet 2 are twice those of planet 1.

What is the free-fall acceleration on planet 2? Express your answer using two significant figures. m/s^2.

Respuesta :

MathPhys

Answer:

13 m/s^2

Explanation:

The acceleration of gravity near the surface of a planet is:

g = MG / R^2

For planet 1, g = 26 m/s^2.

The gravity on planet 2 in terms of the mass and radius of planet 1 is:

g = (2M)G / (2R^2)

g = 1/2 MG / R^2

Since MG/R^2 = 26 m/s^2, then:

g = 13 m/s^2

Cricetus

The free-fall acceleration on planet 2 is "13 m/s²".

According to the question,

→ [tex]g_1 = \frac{GM_1}{R_1^2}[/tex]

       [tex]= 26 \ m/s^2[/tex]

→ [tex]g_2 = \frac{GM_2}{R_2^2}[/tex]

  • [tex]M_2 = 2M_1[/tex]
  • [tex]R_2 = 2R_1[/tex]

Now,

→ [tex]g_2 = \frac{G(2M_1)}{(2R_1)^2}[/tex]

       [tex]= \frac{2GM_1}{4R_1^2}[/tex]

       [tex]= \frac{1}{2} (\frac{GM_1}{R^2} )[/tex]

       [tex]= \frac{1}{2} g_1[/tex]

       [tex]= \frac{1}{2}\times 26[/tex]

       [tex]= 13 \ m/s^2[/tex]

Thus the above answer is right.

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