Question

neptune has a mass that is about 17 times the mass of earth. the distance between the sun and neptune is about 30.1 times the distance between the sun and earth.

if the gravitational force between the sun and earth is 3.5 x 10²⁸ n, which is closest to the force between neptune and the sun?

○ 6 x 10²⁸ n
○ 6 x 10²⁷ n
○ 6 x 10²⁶ n
○ 6 x 10²⁵ n

Explanation:

Step1: Recall Gravitational Force Formula

The gravitational force formula is $F = G\frac{Mm}{r^{2}}$, where $G$ is the gravitational constant, $M$ and $m$ are the masses of the two objects, and $r$ is the distance between their centers. Let the mass of the Sun be $M_{S}$, the mass of Earth be $m_{E}$, and the distance between the Sun and Earth be $r_{E}$. Then the gravitational force between the Sun and Earth is $F_{E}=G\frac{M_{S}m_{E}}{r_{E}^{2}} = 3.5\times 10^{28}\ \text{N}$.

For Neptune, let its mass be $m_{N}=17m_{E}$ and the distance between the Sun and Neptune be $r_{N} = 30.1r_{E}$. The gravitational force between the Sun and Neptune is $F_{N}=G\frac{M_{S}m_{N}}{r_{N}^{2}}$.

Step2: Substitute $m_{N}$ and $r_{N}$ into the Formula

Substitute $m_{N}=17m_{E}$ and $r_{N}=30.1r_{E}$ into the formula for $F_{N}$:
\[

$$\begin{align*} F_{N}&=G\frac{M_{S}(17m_{E})}{(30.1r_{E})^{2}}\\ &=17\times\frac{1}{(30.1)^{2}}\times G\frac{M_{S}m_{E}}{r_{E}^{2}} \end{align*}$$

\]
We know that $G\frac{M_{S}m_{E}}{r_{E}^{2}}=F_{E} = 3.5\times 10^{28}\ \text{N}$, so:
\[
F_{N}=17\times\frac{1}{(30.1)^{2}}\times3.5\times 10^{28}
\]

Step3: Calculate the Numerical Value

First, calculate $\frac{17}{(30.1)^{2}}$:
$(30.1)^{2}\approx30^{2} = 900$ (approximating for simplicity), so $\frac{17}{900}\approx0.0189$.
Then multiply by $3.5\times 10^{28}$:
$0.0189\times3.5\times 10^{28}\approx0.06615\times 10^{28}=6.615\times 10^{26}\ \text{N}$, which is closest to $6\times 10^{26}\ \text{N}$.

Answer:

$6 \times 10^{26}\ \text{N}$ (corresponding to the option "6 x 10²⁶ N")