Physics, 15.10.2019 11:20, fyegirl12

Every few hundred years most of the planets line up on the same side of the sun.(figure 1)calculate the total force on the earth due to venus, jupiter, and saturn, assuming all four planets are in a line, as shown in the figure. the masses are mv=0.815me, mj=318me, msat=95.1me, and the mean distances of the four planets from the sun are 108, 150, 778, and 1430 million km.

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answered: deannajd03

The net Gravitational force acting on the Earth $\fbox{\begin\\9.559 \times {10^{17}}{\text{ N}}\end{minispace}}$ .

Further Explanation:

The net force acting on a body is the vector sum of all the forces acting on the body.

Given:

Venus mass is 0.815M_E .

Jupiter mass is 318 M_E .

Saturn mass is 95.1M_E .

Distance between Venus and Sun is $108\text{ million}\text{ km}$ .

Distance between Earth and Sun is $150\text{ million km}$ .

Distance between Jupiter and Sun is $778\text{ million km}$ .

Distance between Saturn and Sun is $1430\text{ million km}$ .

Concept:

The expression for the Gravitational force of attraction is:

$F = \frac{{G{M_1}{M_2}}}{{{R^2}}}$ …… (1)

The value of the gravitational constant is $6.67 \times {10^{ - 11}}{\text{ }}\frac{{{\text{Nm}}}}{{{\text{k}}{{\text{g}}^2}}}$ .

The mass of the Earth is $5.97 \times {10^{24}}{\text{ kg}}$ .

The expression for the distance between the Earth from the Venus is:

${R_{{\text{EV}}}} = {R_{{\text{SE}}}} - {R_{{\text{SV}}}}$

Substitute 108 for ${R_{{\text{SV}}}}$ and 150 for ${R_{{\text{SE}}}}$ in the above equation.

$\begin{aligned}{R_{{\text{EV}}}}&=\left({150 - 108}\right){\text{ million km}}\\&=42{\text{ million km}}\\&=42\times{10^6}{\text{ km}}\\\end{aligned}$

The expression for the distance between the Earth from the Jupiter is:

${R_{{\text{EJ}}}} = {R_{{\text{SJ}}}} - {R_{{\text{SE}}}}$

Substitute values of ${R_{{\text{SJ}}}}$ and ${R_{{\text{SE}}}}$ .

$\begin{aligned}{R_{{\text{EJ}}}}&=({778 - 150})\\&=628\text{ million km}\\&=628\times{10^6}{\text{ km}}\\\end{aligned}$

The expression for the distance between the Earth from the Saturn is:

${R_{{\text{ESa}}}} = {R_{{\text{SSa}}}} - {R_{{\text{SE}}}}$

Substitute ${R_{{\text{SSa}}}}$ and ${R_{{\text{SE}}}}$ in the above equation.

$\begin{aligned}{R_{{\text{ESa}}}}&=({1430 - 150}){\text{ million km}}\\&=1280{\text{ million km}}\\&=1280\times {10^6}{\text{ km}}\\\end{aligned}$

Substitute ${F_1}$ for F, $42 \times {10^6}{\text{ km}}$ for R and the values of ${M_1}$ , ${M_2}$ , M_E and G in equation (1).

${F_1} =\dfrac{{( {6.67 \times {{10}^{ - 11}}})( {5.97 \times {{10}^{24}}})( {0.815 \times 5.97 \times {{10}^{24}}})}}{{{{( {42 \times {{10}^6{\,km}}})}^2}}}$

$\begin{aligned} {F_1}&=\frac{{( {6.67 \times {{10}^{ - 11}}})( {5.97 \times {{10}^{24}}})( {0.815 \times 5.97 \times {{10}^{24}}})}}{{{{( {42 \times {{10}^9}}} )}^2}}}\\{F_1}&= 0.1098 \times {10^{19}}{\text{ N}}\end{aligned}$

Substitute ${F_2}$ for F, $628 \times {10^6}{\text{ km}}$ for R and values of ${M_1}$ , ${M_2}$ , M_E and G in equation (1).

${F_2}=\dfrac{{( {6.67 \times {{10}^{ - 11}}})( {5.97 \times {{10}^{24}}})( {318 \times 5.97 \times {{10}^{24}}})}}{{{{( {628 \times {{10}^6{\,km}}})}^2}}}$

$\begin{aligned} {F_2}&=\frac{{( {6.67 \times {{10}^{ - 11}}})( {5.97 \times {{10}^{24}}})({318 \times 5.97 \times {{10}^{24}}})}}{{{{( {628 \times {{10}^9}})}^2}}}\\{F_2}&=0.1916 \times {10^{19}}{\text{ N}}\\\end{aligned}$

Substitute ${F_3}$ for F, $1280 \times {10^6}{\text{ km}}$ for R and the values of ${M_1}$ , ${M_2}$ , M_E and G in equation (1).

$\begin{aligned} {F_3}&= \frac{{( {6.67 \times {{10}^{ - 11}}})( {5.97 \times {{10}^{24}}})( {95.1 \times 5.97 \times {{10}^{24}}})}}{{{{( {1280 \times {{10}^9{\,m}}} )}^2}}}\\ {F_3}&= 0.1379 \times {10^{18}}{\text{ N}}\end{aligned}$

The force acting on the left side is taken negative and force acting on the left side is taken positive as shown in the redrawn figure.

The expression net force acting on the Earth is:

$\fbox{\begin\\F = - {F_1} + {F_2} + {F_3}\end{minispace}}$

Here, F is the net Gravitational force acting on the Earth.

Substitute the values in the above equation.

$\begin{aligned} F&=-0.1098 \times {10^{19}}+0.1916 \times {10^{19}}+0.1379 \times {10^{18}}\\&= 9.559 \times {10^{17}}{\text{ N}} \\ \end{aligned}$

Therefore, the net Gravitational force acting on the Earth $\fbox{\begin\\9.559 \times {10^{17}}{\text{ N}}\end{minispace}}$ .

Learn more:

1. Motion of ball under gravity

2. The motion of a body under friction

3. Net force on a body

Answer Details:

Grade: College

Subject: Physics

Chapter: Gravitation

Keywords:

Every few hundred years, planets, line up, same side, Sun, the Venus, Earth, Jupiter and Saturn, masses, MV=0.815ME, MJ=318 ME and MSat=95.1 ME, mean distance, four planets, 108, 150, 778, and 1430 million km, 9.559x 10^17 N.