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(a) Calculate little g on the surface of the Sun. (b) Now, say 90 kg B is in an elevator on the Sun with the international scale and the elevator is accelerating down at 25 ge , where ge = 10 m/s 2 . What does the scale display? Use G = 6.67 ⨉ 10 -11 N m2 /kg 2 , Sun mass: 2 ⨉ 10 30 kg, Sun radius: 700,000 km.

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Answer

a) to find the acceleration due to gravity of sun

       [tex]g = \dfrac{GM}{R^2}[/tex]

m is mass of sun = 2 x 10³⁰ Kg

    R is the radius =  700,000 km = 7 x 10⁷ m

       [tex]g = \dfrac{6.67\times 10^{-11}\times 2 \times 10^{30}}{(7\times 10^7)^2}[/tex]

             g = 272.24 m/s²

b) mass of the man = 90 Kg

  acceleration = 25 g

   [tex]F_{net} = m (g_{sun} - 25 g)[/tex]

   [tex]F_{net} = 90\times (272.24 - 25\times 10)[/tex]

   [tex]F_{net} =2001.6\ N[/tex]

reading of scale on sun

   [tex]m' = \dfrac{F_{net}}{g_{sun}}[/tex]

   [tex]m' = \dfrac{2001.6}{272.24}[/tex]

        m' = 7.35 Kg

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