Question #e1c2c

1 Answer
May 28, 2016

Answer:

-1.2%

Explanation:

Let

  • M be the mass of Mars
  • R be the radius of Mars
  • D be the distance of the top point of the highest mountain from the center of Mars
  • The gravitational field strength at the bottom of the mountain or at the surface of the mars
    # g_b=(GM)/R^2........(1)# ,Where G = Gravitational constant

  • Similarly the gravitational field strength at the top of the mountain .

  • # g_t=(GM)/D^2........(2)#

By the condition of the problem

#D = R+0.6%of R=R(1+0.006)#

Dividing equation(2) by equation (1) we get

  • #g_t/(g _b) = (cancelR/(cancelR(1+0.006)))^2=(1+.006)^-2#

#g_t/(g _b) ~~1-0.012# (Neglecting higher power values)

The change in gravitational field strength from the bottom to the top of Olympus Mons is

#(g_t-g_b)/(g _b) =-0.012=-1.2%#

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