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2. ceres, the largest asteroid known, has a mass of roughly 8.7×10^20 k…

Question

  1. ceres, the largest asteroid known, has a mass of roughly 8.7×10^20 kg. if ceres passes within 14,000 km of the spaceship in which you and your dog are traveling, what force does it exert on your dog? the mass of your dog is 30 kg. (treat your dog and the asteroid as point objects.)

Explanation:

Step1: Convert distance to SI unit

The distance $r = 14000\ km=14000\times10^{3}\ m = 1.4\times 10^{7}\ m$. The mass of Ceres $M = 8.7\times 10^{20}\ kg$ and the mass of the dog $m = 30\ kg$. The gravitational - constant $G=6.67\times 10^{- 11}\ N\cdot m^{2}/kg^{2}$.

Step2: Apply Newton's law of universal gravitation

The formula for the gravitational force between two objects is $F = G\frac{Mm}{r^{2}}$. Substitute the values: $F=(6.67\times 10^{-11}\ N\cdot m^{2}/kg^{2})\frac{(8.7\times 10^{20}\ kg)\times(30\ kg)}{(1.4\times 10^{7}\ m)^{2}}$.
First, calculate the numerator: $(6.67\times 10^{-11})\times(8.7\times 10^{20})\times30=6.67\times8.7\times30\times10^{-11 + 20}=6.67\times8.7\times30\times10^{9}$.
$6.67\times8.7\times30=(6.67\times8.7)\times30 = 58.029\times30 = 1740.87$. So the numerator is $1740.87\times10^{9}\ N\cdot m^{2}/kg$.
The denominator is $(1.4\times 10^{7})^{2}=1.4^{2}\times10^{14}=1.96\times 10^{14}\ m^{2}$.
Then $F=\frac{1740.87\times 10^{9}}{1.96\times 10^{14}}\ N$.
$\frac{1740.87}{1.96}\approx888.2$. And $\frac{10^{9}}{10^{14}}=10^{9 - 14}=10^{-5}$.
So $F = 888.2\times10^{-5}\ N\approx0.089\ N$.

Answer:

$0.089\ N$