Question

the mass of the sun is 1.99×10³⁰ kg. jupiter is 7.79×10⁸ km away from the sun and has a mass of 1.90×10²⁷ kg. the gravitational force between the sun and jupiter to three significant figures is ×10²³ n. 2.86 3.24 4.16 5.44

Explanation:

Step1: Recall gravitational - force formula

The gravitational - force formula is $F = G\frac{m_1m_2}{r^{2}}$, where $G = 6.67\times10^{- 11}\text{ N}\cdot\text{m}^{2}/\text{kg}^{2}$, $m_1$ and $m_2$ are the masses of the two objects, and $r$ is the distance between them. First, convert the distance from kilometers to meters. Given $r = 7.79\times10^{8}\text{ km}=7.79\times10^{11}\text{ m}$, $m_1 = 1.99\times10^{30}\text{ kg}$, and $m_2 = 1.90\times10^{27}\text{ kg}$.

Step2: Substitute values into the formula

$F=6.67\times 10^{-11}\frac{1.99\times 10^{30}\times1.90\times 10^{27}}{(7.79\times 10^{11})^{2}}$.
First, calculate the numerator: $1.99\times10^{30}\times1.90\times10^{27}=(1.99\times1.90)\times10^{30 + 27}=3.781\times10^{57}$.
Then, calculate the denominator: $(7.79\times10^{11})^{2}=7.79^{2}\times10^{22}=60.6841\times10^{22}=6.06841\times10^{23}$.
Now, $F = 6.67\times10^{-11}\times\frac{3.781\times10^{57}}{6.06841\times10^{23}}$.
Using the rule of exponents $\frac{a\times10^{m}}{b\times10^{n}}=\frac{a}{b}\times10^{m - n}$, we have $\frac{3.781\times10^{57}}{6.06841\times10^{23}}\approx0.623\times10^{34}=6.23\times10^{33}$.
Then $F = 6.67\times10^{-11}\times6.23\times10^{33}$.
Using the rule of exponents $a\times10^{m}\times b\times10^{n}=ab\times10^{m + n}$, we get $F=(6.67\times6.23)\times10^{-11 + 33}=41.5541\times10^{22}\approx4.16\times10^{23}\text{ N}$.

Answer:

4.16×10²³ N