8. neptune orbits the sun at an average distance of ( 4.495 imes10^{12}m ). if the mass of the sun is ( 1.99 imes10^{30}kg ), and the mass of neptune is ( 1.024 imes10^{26} ), what is the gravitational force between them? ( 4.53cdot10^{43}n ) ( 6.70cdot10^{20}n ) ( 1.01cdot10^{31}n ) ( 3.02cdot10^{33}n ) clear all

Question

8. neptune orbits the sun at an average distance of ( 4.495	imes10^{12}m ). if the mass of the sun is ( 1.99	imes10^{30}kg ), and the mass of neptune is ( 1.024	imes10^{26} ), what is the gravitational force between them? ( 4.53cdot10^{43}n ) ( 6.70cdot10^{20}n ) ( 1.01cdot10^{31}n ) ( 3.02cdot10^{33}n ) clear all

Accepted Answer

# Explanation:
## Step1: Recall the gravitational force formula
The formula for gravitational force is $F = G\frac{m_1m_2}{r^2}$, where $G = 6.67\times10^{- 11}\space N\cdot m^{2}/kg^{2}$, $m_1$ is the mass of the Sun ($m_1=1.99\times 10^{30}\space kg$), $m_2$ is the mass of Neptune ($m_2 = 1.024\times10^{26}\space kg$), and $r$ is the average distance ($r = 4.495\times10^{12}\space m$).
## Step2: Substitute the values into the formula
$$
\begin{align*}
F&=(6.67\times 10^{-11})\frac{(1.99\times 10^{30})(1.024\times 10^{26})}{(4.495\times 10^{12})^{2}}\
&=(6.67\times 10^{-11})\frac{1.99\times1.024\times10^{30 + 26}}{4.495^{2}\times10^{24}}\
&=(6.67\times 10^{-11})\frac{2.03776\times10^{56}}{20.205025\times10^{24}}\
&=(6.67\times 10^{-11})\times(0.10085\times10^{32})\
&=6.67\times0.10085\times10^{-11 + 32}\
&=0.6727\times10^{21}\
&=6.727\times 10^{20}\space N
\end{align*}
$$

# Answer:
$6.70\times 10^{20}\space N$ (the small difference is due to rounding in intermediate steps)

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Question

Explanation:

Step1: Recall the gravitational force formula

Step2: Substitute the values into the formula

Answer: