Question

21-22) develop a mathematical model to calculate the gravitational force between mars and phobos. show all work, including an equation and substitution with units. 2
$f_g = \frac{gm_1m_2}{r^2}$
$f_g = \frac{(6.67\times10^{-11} n\cdot m^2/kg^2)(6.42\times10^{23} kg)(1.07\times10^{16} kg)}{(9.38\times10^6 m)^2}$
$m_1 = mars$
$m_2 = phobos$
$g = earth$

Explanation:

Step1: Recall the gravitational force formula

The formula for gravitational force between two objects is \( F_g=\frac{Gm_1m_2}{r^2} \), where \( G = 6.67\times 10^{-11}\space N\cdot m^2/kg^2 \) (universal gravitational constant), \( m_1 \) and \( m_2 \) are the masses of the two objects, and \( r \) is the distance between their centers. Here, \( m_1 = 6.42\times 10^{23}\space kg \) (mass of Mars), \( m_2=1.07\times 10^{16}\space kg \) (mass of Phobos), and \( r = 9.38\times 10^{6}\space m \) (distance between Mars and Phobos).

Step2: Substitute the values into the formula

Substitute \( G = 6.67\times 10^{-11}\space N\cdot m^2/kg^2 \), \( m_1 = 6.42\times 10^{23}\space kg \), \( m_2 = 1.07\times 10^{16}\space kg \), and \( r = 9.38\times 10^{6}\space m \) into the formula:

\[

$$\begin{align*} F_g&=\frac{(6.67\times 10^{-11}\space N\cdot m^2/kg^2)\times(6.42\times 10^{23}\space kg)\times(1.07\times 10^{16}\space kg)}{(9.38\times 10^{6}\space m)^2}\\ \end{align*}$$

\]

First, calculate the numerator: \((6.67\times 10^{-11})\times(6.42\times 10^{23})\times(1.07\times 10^{16})\)

\[

$$\begin{align*} &(6.67\times 6.42\times 1.07)\times(10^{-11 + 23+ 16})\\ &\approx(6.67\times 6.42\times 1.07)\times10^{28}\\ &6.67\times 6.42\approx 42.8214\\ &42.8214\times 1.07\approx 45.8189\\ &\text{Numerator}\approx 45.8189\times 10^{28}=4.58189\times 10^{29} \end{align*}$$

\]

Then, calculate the denominator: \((9.38\times 10^{6})^2=(9.38)^2\times(10^{6})^2 = 88.0644\times 10^{12}=8.80644\times 10^{13}\)

Step3: Divide numerator by denominator

\[
F_g=\frac{4.58189\times 10^{29}}{8.80644\times 10^{13}}\approx\frac{4.58189}{8.80644}\times 10^{29 - 13}
\]

\[
\frac{4.58189}{8.80644}\approx 0.5203
\]

\[
F_g\approx 0.5203\times 10^{16}=5.203\times 10^{15}\space N
\]

Answer:

The gravitational force between Mars and Phobos is approximately \( \boldsymbol{5.2\times 10^{15}\space N} \) (rounded to a reasonable number of significant figures).