# Question e1c2c

May 28, 2016

-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
${g}_{b} = \frac{G M}{R} ^ 2. \ldots \ldots . \left(1\right)$ ,Where G = Gravitational constant

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

• ${g}_{t} = \frac{G M}{D} ^ 2. \ldots \ldots . \left(2\right)$

By the condition of the problem

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

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

• ${g}_{t} / \left({g}_{b}\right) = {\left(\frac{\cancel{R}}{\cancel{R} \left(1 + 0.006\right)}\right)}^{2} = {\left(1 + .006\right)}^{-} 2$

${g}_{t} / \left({g}_{b}\right) \approx 1 - 0.012$ (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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