Question

a planet with a mass of 4.5 x 10^11 kg is orbiting a star with a mass of 6.0 x 10^17 kg. the distance from their centers is 7.9 x 10^10 m. what is the gravitational force between them?

Explanation:

Step1: Identify the formula

The gravitational - force formula is $F = G\frac{m_1m_2}{r^2}$, where $G = 6.67\times10^{- 11}\ Nm^2/kg^2$, $m_1$ and $m_2$ are the masses of the two objects, and $r$ is the distance between their centers.

Step2: Substitute the given values

Let $m_1 = 4.5\times10^{11}\ kg$, $m_2 = 6.0\times10^{17}\ kg$, and $r = 7.9\times10^{10}\ m$. Then $F=(6.67\times10^{-11})\frac{(4.5\times10^{11})(6.0\times10^{17})}{(7.9\times10^{10})^2}$.

Step3: Calculate the numerator

$(4.5\times10^{11})(6.0\times10^{17})=4.5\times6.0\times10^{11 + 17}=27\times10^{28}=2.7\times10^{29}$.

Step4: Calculate the denominator

$(7.9\times10^{10})^2 = 7.9^2\times10^{20}=62.41\times10^{20}=6.241\times10^{21}$.

Step5: Calculate the force

$F=(6.67\times10^{-11})\frac{2.7\times10^{29}}{6.241\times10^{21}}$. First, $\frac{2.7\times10^{29}}{6.241\times10^{21}}=\frac{2.7}{6.241}\times10^{29 - 21}\approx0.4326\times10^{8}=4.326\times10^{7}$. Then $F=(6.67\times10^{-11})\times(4.326\times10^{7})=6.67\times4.326\times10^{-11 + 7}\approx28.85\times10^{-4}=2.89\times10^{-3}\ N$.

Answer:

$2.89\times10^{-3}\ N$