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Q6.3-34 PE

College Physics (Urone)
Found in: Page 223

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Short Answer

(a) Calculate the magnitude of the acceleration due to gravity on the surface of Earth due to the Moon. (b) Calculate the magnitude of the acceleration due to gravity at Earth due to the Sun. (c) Take the ratio of the Moon’s acceleration to the Sun’s and comment on why the tides are predominantly due to the Moon in spite of this number.

(a) Magnitude of acceleration to the moon, \({{\rm{a}}_{\rm{m}}}{\rm{ = 3}}{\rm{.43x1}}{{\rm{0}}^{{\rm{ - 5}}}}{\rm{ m/}}{{\rm{s}}^{\rm{2}}}\).

(b) Magnitude of acceleration to the sun, \({{\rm{a}}_{\rm{s}}}{\rm{ = 5}}{\rm{.93x1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{ m/}}{{\rm{s}}^{\rm{2}}}\) .

(c) Ratio of both acceleration of moon to sun\(\frac{{{{\rm{a}}_{\rm{m}}}}}{{{{\rm{a}}_{\rm{s}}}}}{\rm{ = 5}}{\rm{.78x1}}{{\rm{0}}^{{\rm{ - 3}}}}\).

See the step by step solution

Step by Step Solution

Step 1: Definition of Gravity

Gravity is a universal phenomenon and is introduced by Newton and Derived the expression for gravitational force.

Step 2: Given Data

Mass of the earth \({{\rm{M}}_{\rm{e}}}{\rm{ = 5}}{\rm{.979 x 1}}{{\rm{0}}^{{\rm{24}}}}{\rm{ kg}}\)

Mass of the moon \({{\rm{M}}_{\rm{m}}}{\rm{ = 7}}{\rm{.3477 x 1}}{{\rm{0}}^{{\rm{22}}}}{\rm{ kg}}\)

Mass of the sun \({{\rm{M}}_{\rm{s}}}{\rm{ = 1}}{\rm{.9891 x 1}}{{\rm{0}}^{{\rm{30}}}}{\rm{ kg}}\)

Distance from surface of the earth to moon

\(\begin{aligned}{}{{\rm{R}}_{\rm{m}}}{\rm{ = }}\left( {{\rm{3}}{\rm{.84 x 1}}{{\rm{0}}^{\rm{8}}}{\rm{-- 6371 x1}}{{\rm{0}}^{\rm{3}}}} \right)\\{{\rm{R}}_{\rm{m}}}{\rm{ = 3}}{\rm{.78 x 1}}{{\rm{0}}^{\rm{8}}}{\rm{ m}}\end{aligned}\)

Distance from surface of the earth to sun

\(\begin{aligned}{}{{\rm{R}}_{\rm{s}}}{\rm{ = }}\left( {{\rm{1}}{\rm{.496 x 1}}{{\rm{0}}^{{\rm{11}}}}{\rm{-- 6371 x 1}}{{\rm{0}}^{\rm{3}}}} \right)\\{{\rm{R}}_{\rm{s}}}{\rm{ = 1}}{\rm{.496 x 1}}{{\rm{0}}^{{\rm{11}}}}{\rm{ m}}\end{aligned}\)

Acceleration due to gravity due to moon \({{\rm{a}}_{\rm{m}}}{\rm{ = ?}}\)

Acceleration due to gravity due to sun \({{\rm{a}}_{\rm{s}}}{\rm{ = ?}}\)

Step 3: Calculation of Magnitude of acceleration to moon

The gravitational force on the surface of the earth due to the moon is given by the equation

\({\rm{F = G}}\frac{{{{\rm{M}}_{\rm{e}}}{{\rm{M}}_{\rm{m}}}}}{{{{\rm{R}}_{\rm{m}}}^{\rm{2}}}}\)

Substituting the value of force in this equation, we get

\({{\rm{M}}_{\rm{e}}}{{\rm{a}}_{\rm{m}}}{\rm{ = G}}\frac{{{{\rm{M}}_{\rm{e}}}{{\rm{M}}_{\rm{m}}}}}{{{{\rm{R}}_{\rm{m}}}^{\rm{2}}}}\)

\({{\rm{a}}_{\rm{m}}}{\rm{ = G}}\frac{{{{\rm{M}}_{\rm{m}}}}}{{{{\rm{R}}_{\rm{m}}}^{\rm{2}}}}\)

\({{\rm{a}}_{\rm{m}}}{\rm{ = 6}}{\rm{.673 \times 1}}{{\rm{0}}^{{\rm{ - 11}}}}\left( {\frac{{{\rm{7}}{\rm{.3477 \times 1}}{{\rm{0}}^{{\rm{22}}}}}}{{{{{\rm{(3}}{\rm{.78x1}}{{\rm{0}}^{\rm{8}}}{\rm{)}}}^{\rm{2}}}}}} \right)\)

\({{\rm{a}}_{\rm{m}}}{\rm{ = 3}}{\rm{.43x1}}{{\rm{0}}^{{\rm{ - 5}}}}{\rm{ m/}}{{\rm{s}}^{\rm{2}}}\)

Step 4: Calculation of Magnitude of acceleration to sun

Similarly, the gravitational force exerted by the sun on the earth's surface is determined by the equation.

\(\begin{aligned}{}{{\rm{M}}_{\rm{e}}}{{\rm{a}}_{\rm{s}}}{\rm{ = G}}\frac{{{{\rm{M}}_{\rm{e}}}{{\rm{M}}_{\rm{s}}}}}{{{{\rm{R}}_{\rm{s}}}^{\rm{2}}}}\\{{\rm{a}}_{\rm{s}}}{\rm{ = G}}\frac{{{{\rm{M}}_{\rm{s}}}}}{{{{\rm{R}}_{\rm{s}}}^{\rm{2}}}}\end{aligned}\)

\({{\rm{a}}_{\rm{s}}}{\rm{ = 6}}{\rm{.673 \times 1}}{{\rm{0}}^{{\rm{ - 11}}}} \times \left( {\frac{{{\rm{1}}{\rm{.9891 \times 1}}{{\rm{0}}^{{\rm{30}}}}}}{{{{{\rm{(1}}{\rm{.496x1}}{{\rm{0}}^{{\rm{11}}}}{\rm{)}}}^{\rm{2}}}}}} \right)\)

\({{\rm{a}}_{\rm{s}}}{\rm{ = 5}}{\rm{.93x1}}{{\rm{0}}^{{\rm{ - 3}}}}{\rm{ m/}}{{\rm{s}}^{\rm{2}}}\)

Step 5: Calculation of ratio of both acceleration of moon to sun

The ratio of the moon’s acceleration to the sun’s acceleration

\(\begin{aligned}{}\frac{{{{\rm{a}}_{\rm{m}}}}}{{{{\rm{a}}_{\rm{s}}}}}{\rm{ = }}\frac{{{\rm{3}}{\rm{.43x1}}{{\rm{0}}^{{\rm{ - 5}}}}}}{{{\rm{5}}{\rm{.93x1}}{{\rm{0}}^{{\rm{ - 3}}}}}}\\\frac{{{{\rm{a}}_{\rm{m}}}}}{{{{\rm{a}}_{\rm{s}}}}}{\rm{ = 5}}{\rm{.78x1}}{{\rm{0}}^{{\rm{ - 3}}}}\end{aligned}\)

If we take the difference of centripetal acceleration on two opposite sides of the moon, we get a difference of \({\rm{2}}{\rm{.2 x 1}}{{\rm{0}}^{{\rm{ - 6}}}}{\rm{ m/}}{{\rm{s}}^{\rm{2}}}\) and for two opposite sides of the sun, this difference is \({\rm{1 x 1}}{{\rm{0}}^{{\rm{ - 6}}}}{\rm{ m/}}{{\rm{s}}^{\rm{2}}}\). This difference tells us Centripetal force on earth is more influenced by the moon rather than the sun.

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