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The existence of the planet Pluto was proposed based on irregularities in Neptune's orbit. Pluto...

Question:

The existence of the planet Pluto was proposed based on irregularities in Neptune's orbit. Pluto was subsequently discovered near its predicted position. But it now appears that the discovery was fortuitous, since Pluto is small and the irregularities in Neptune's orbit were not well known. Answer the following to illustrate that Pluto has a minor effect on the orbit of Neptune compared with other planets:

a) Calculate the acceleration of gravity at Neptune due to Pluto when they are {eq}5.00 \times 10^{12} \ m {/eq} apart. The mass of Pluto is {eq}1.4 \times 10^{22} \ kg {/eq}

b) Calculate the acceleration of gravity at Neptune due to Uranus, when they are {eq}2.50 \times 10^{12} \ m {/eq} apart. The mass of Uranus is {eq}8.62 \times 10^{25} \ kg {/eq}.

c. Give the ratio of this acceleration to that due to Pluto.

Gravitational Force:

The value of gravitational force can be evaluated by examining the gravitational constant, the gap between the two objects, the mass of the first object, and the mass of the second object. It varies inversely to the value of the distance between the two objects.

Answer and Explanation:


Part a)


Given data:

  • Mass of Pluto, {eq}{m_P} = 1.4 \times {10^{22}}\;{\rm{kg}} {/eq}
  • Distance between Neptune and Pluto, {eq}{r_1} = 5.00 \times {10^{12}}\;{\rm{m}} {/eq}


The expression for the gravitational force is,

{eq}F = \dfrac{{G{m_P}{m_N}}}{{r_1^2}} {/eq}

The expression for the force is,

{eq}\begin{align*} F &= {m_N}{a_1}\\ {a_1} &= \dfrac{F}{{{m_N}}} \end{align*} {/eq}

Substitute the value of equation (1) in above equation

{eq}\begin{align*} {a_1} &= \dfrac{{G{m_P}{m_N}}}{{{m_N}r_1^2}}\\ {a_1} &= \dfrac{{G{m_P}}}{{r_1^2}} \end{align*} {/eq}

Here, {eq}G = 6.67 \times {10^{ - 11}}\;{\rm{N}} \cdot {{\rm{m}}^{\rm{2}}}{\rm{/k}}{{\rm{g}}^{\rm{2}}} {/eq} is the gravitational constant.

Substitute the given values,

{eq}\begin{align*} {a_1} &= \dfrac{{6.67 \times {{10}^{ - 11}} \times 1.4 \times {{10}^{22}}}}{{{{\left( {5.00 \times {{10}^{12}}} \right)}^2}}}\\ {a_1} &= 3.73 \times {10^{ - 14}}\;{\rm{m/}}{{\rm{s}}^{\rm{2}}} \end{align*} {/eq}

Therefore, the acceleration of gravity at Neptune due to Pluto is {eq}3.73 \times {10^{ - 14}}\;{\rm{m/}}{{\rm{s}}^{\rm{2}}} {/eq} .


Part b)


Given data:

  • The distance between Neptune and Uranus, {eq}{r_2} = 2.50 \times {10^{12}}\;{\rm{m}} {/eq}
  • Mass of Uranus, {eq}{m_U} = 8.62 \times {10^{25}}\;{\rm{kg}} {/eq}


The acceleration can be calculated as,

{eq}\begin{align*} {a_2} &= \dfrac{{G{m_U}}}{{r_2^2}}\\ {a_2} &= \dfrac{{6.67 \times {{10}^{ - 11}} \times 8.62 \times {{10}^{25}}}}{{{{\left( {2.50 \times {{10}^{12}}} \right)}^2}}}\\ {a_2} &= 9.19 \times {10^{ - 10}}\;{\rm{m/}}{{\rm{s}}^{\rm{2}}} \end{align*} {/eq}

Therefore, the acceleration of gravity at Neptune due to Uranus is {eq}9.19 \times {10^{ - 10}}\;{\rm{m/}}{{\rm{s}}^{\rm{2}}} {/eq} .


Part c)

The ratio of this acceleration to that due to Pluto is,

{eq}\begin{align*} R &= \dfrac{{{a_2}}}{{{a_1}}}\\ R &= \dfrac{{9.19 \times {{10}^{ - 10}}}}{{3.73 \times {{10}^{ - 14}}}}\\ R &= 2.46 \times {10^4} \end{align*} {/eq}

Therefore, the value of ratio is {eq}2.46 \times {10^4} {/eq} .


Learn more about this topic:

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Gravitational Force: Definition, Equation & Examples

from Ohio State Test - Physical Science: Practice & Study Guide

Chapter 13 / Lesson 6
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