Question

(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.

Answer 1

a) to find the acceleration due to gravity of sun

`g = (GM)/(R^2)`

m is mass of sun = 2 x 10³⁰ Kg

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

`g = (6.67* 10^(-11)* 2 * 10^(30))/((7* 10^7)^2)`

g = 272.24 m/s²

b) mass of the man = 90 Kg

acceleration = 25 g

`F_(net) = m (g_(sun) - 25 g)`

`F_(net) = 90* (272.24 - 25* 10)`

`F_(net) =2001.6\ N`

reading of scale on sun

`m' = (F_(net))/(g_(sun))`

`m' = (2001.6)/(272.24)`

m' = 7.35 Kg