Question

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.

Answer 1

Answer: $3.7 \times 10^{-4} N$

Explanation:

The gravitational pull between two object is given by:

$F = G\frac{Mm}{r^2}$

Where M and m are the masses of the object, r is the distance between the masses and G = 6.67× 10⁻¹¹ m³kg⁻¹ s⁻² is the gravitational constant.

We have to calculate the net force on Earth due to Venus, Jupiter and Saturn when they are in one line. It means when they are the closest distance.

$F_{net] = G\frac{M_eM_v}{r_v^2}+G\frac{M_eM_j}{r_j^2}+G\frac{M_eM_s}{r_s^2}$

Mass of Earth, Me = 5.98 × 10²⁴ kg

Mass of Venus, Mv = 0.815 Me

Mass of Jupiter, Mj = 318 Me

Mass of Saturn, Ms = 95.1 Me

closest distance between Earth and Venus, rv = 38 × 10⁶ km = 0.25 AU

closest distance between Jupiter and Earth, rj = 588 × 10⁶ km = 3.93 AU

closest distance between Earth and Saturn, rs = 1.2 × 10⁹ km = 8.0 AU

where 1 AU = 1.5 × 10¹¹ m

Inserting the values:

$F_{net} = G\frac{M_e\times 0.815 M_e}{(0.25AU)^2}+G\frac{M_e\times 318 M_e}{(3.93AU)^2}+G\frac{M_e\times 95.1 M_e}{(8.0AU)^2}\\ \Rightarrow F_{net} = \frac{(GM_e^2)}{(1AU)^2}(\frac{0.815}{0.25^2}+\frac{318}{3.93^2}+\frac{95.1}{8.0^2})=\frac{6.67\times 10^{-11} \times (5.98\times 10^{24})^2}{(1.5\times 10^{11})^2}(35.1) = 3.7 \times 10^{-4} N$

Answer 2

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 brainly.com/question/10934170

2. The motion of a body under friction brainly.com/question/7031524

3. Net force on a body brainly.com/question/4033012

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.