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Q. 4.3

FORCES OF DISTANT WORLDS

GOAL Calculate the magnitude of a gravitational force using Newton’s law of gravitation.

PROBLEM (a) Find the gravitational force exerted by the Sun on a 70.0-kg man located at the Earth’s equator at noon, when the man is closest to the Sun. (b) Calculate the gravitational force of the Sun on the man at midnight, when he is farthest from the Sun. (c) Calculate the difference in the acceleration due to the Sun between noon and midnight. (For values, see Table 7.3 on page 214.)

Table 7.3 Useful planetary Data

$\begin{array}{lccrrr}\textbf{Body} & \textbf{Mass (kg)} & \begin{array}{r} \textbf{mean} \qquad\\ \textbf{Radius (m)}\end{array} & \textbf{Period (s)} & \begin{array}{r} \textbf{mean} \qquad \\ \textbf{Distance} \quad \\ \textbf{from Sun (m)} \end{array} & \frac{\textbf{T}^\textbf{2}}{\textbf{r}^\textbf{3}} \textbf{10}^{\textbf{-19}}\left( \frac{\textbf{s}^\textbf{2}}{\textbf{m}^\textbf{2}} \right) \\ \hline \text { Mercury } & 3.18 \times 10^{23} & 2.43 \times 10^6 & 7.60 \times 10^6 & 5.79 \times 10^{10} & 2.97 \qquad \\ \text { Venus } & 4.88 \times 10^{24} & 6.06 \times 10^6 & 1.94 \times 10^7 & 1.08 \times 10^{11} & 2.99 \qquad \\ \text { Earth } & 5.98 \times 10^{24} & 6.38 \times 10^6 & 3.156 \times 10^7 & 1.496 \times 10^{11} & 2.97 \qquad \\ \text { Mars } & 6.42 \times 10^{23} & 3.37 \times 10^6 & 5.94 \times 10^7 & 2.28 \times 10^{11} & 2.98 \qquad \\ \text { Jupiter } & 1.90 \times 10^{27} & 6.99 \times 10^7 & 3.74 \times 10^8 & 7.78 \times 10^{11} & 2.97 \qquad \\ \text { Saturn } & 5.68 \times 10^{26} & 5.85 \times 10^7 & 9.35 \times 10^8 & 1.43 \times 10^{12} & 2.99 \qquad \\ \text { Uranus } & 8.68 \times 10^{25} & 2.33 \times 10^7 & 2.64 \times 10^9 & 2.87 \times 10^{12} & 2.95 \qquad \\ \text { Neptune } & 1.03 \times 10^{26} & 2.21 \times 10^7 & 5.22 \times 10^9 & 4.50 \times 10^{12} & 2.99 \qquad \\ \text { Pluto }^{\mathrm{a}} & 1.27 \times 10^{23} & 1.14 \times 10^6 & 7.82 \times 10^9 & 5.91 \times 10^{12} & 2.96 \qquad \\ \text { Moon } & 7.36 \times 10^{22} & 1.74 \times 10^6 & – & – & – \qquad \\ \text { Sun } & 1.991 \times 10^{30} & 6.96 \times 10^8 & – & – & – \qquad \end{array}$

$^{\mathrm{a}}$ In August 2006, the International Astronomical Union adopted a definition of a planet that separates Pluto from the other eight planets. Pluto is now defined as a “dwarf planet” like the asteroid Ceres.

STRATEGY To obtain the distance of the Sun from the man at noon, subtract the Earth’s radius from the solar distance. At midnight, add the Earth’s radius. Retain enough digits so that rounding doesn’t remove the small difference between the two answers. For part (c), subtract the answer for (b) from (a) and divide by the man’s mass.

Verified Solution

(a) Find the gravitational force exerted by the Sun on the man at the Earth’s equator at noon.

Write the law of gravitation, Equation $4.5$,

$F_g = G \frac{m_1 m_2}{r^2}$    [4.5]

in terms of the distance from the Sun to the Earth, $r_{S}$, and Earth’s radius, $R_{E}$ :

(1) $\quad F_{\text {Sun }}^{\text {noon }}=\frac{m M_{S} G}{r^{2}}=\frac{m M_{S} G}{\left(r_{S} – R_{E}\right)^{2}}$

Substitute values into (1) and retain two extra digits:

\begin{aligned}F_{\text {Sun }}^{\text {noon }} &=\frac{(70.0 \mathrm{~kg})\left(1.991 \times 10^{30} \mathrm{~kg}\right)\left(6.67 \times 10^{-11} \mathrm{~kg}{ }^{-1} \mathrm{~m}^{3} / \mathrm{s}^{2}\right)}{\left(1.496 \times 10^{11} \mathrm{~m} – 6.38 \times 10^{6} \mathrm{~m}\right)^{2}} \\&=0.415 40 \mathrm{~N}\end{aligned}

(b) Calculate the gravitational force of the Sun on the man at midnight. Write the law of gravitation, adding Earth’s radius this time:

(2) $F_{\text {Sun }}^{\text {mid }}=\frac{m M_{S} G}{r^{2}}=\frac{m M_{S} G}{\left(r_{S}+R_{E}\right)^{2}}$

Substitute values into (2):

\begin{aligned}F_{\text {Sun }}^{\text {mid }} &=\frac{(70.0 \mathrm{~kg})\left(1.991 \times 10^{30} \mathrm{~kg}\right)\left(6.67 \times 10^{-11} \mathrm{~kg}{ }^{-1} \mathrm{~m}^{3} / \mathrm{s}^{2}\right)}{\left(1.496 \times 10^{11} \mathrm{~m}+6.38 \times 10^{6} \mathrm{~m}\right)^{2}} \\&=0.415 33 \mathrm{~N}\end{aligned}

(c) Calculate the difference in the man’s solar acceleration between noon and midnight.

Write an expression for the difference in acceleration and substitute values:

\begin{aligned}a &=\frac{F_{\text {Sun }}^{\text {noon }}-F_{\text {Sun }}^{\text {mid }}}{m}=\frac{0.415 40 \mathrm{~N} – 0.415 33 \mathrm{~N}}{70.0 \mathrm{~kg}} \\& \cong 1 \times 10^{-6} \mathrm{~m} / \mathrm{s}^{2}\end{aligned}

REMARKS The gravitational attraction between the Sun and objects on Earth is easily measurable and has been exploited in experiments to determine whether gravitational attraction depends on the composition of the object. The gravitational force on Earth due to the Moon is much weaker than the gravitational force on Earth due to the Sun. Paradoxically, the Moon’s effect on the tides is over twice that of the Sun because the tides depend on differences in the gravitational force across the Earth, and those differences are greater for the Moon’s gravitational force because the Moon is much closer to Earth than the Sun.