Sure! Let's break down the steps needed to solve this problem. We are given the following values:
- Gravitational force ([tex]\( F \)[/tex]) = [tex]\( 2.58 \times 10^3 \)[/tex] N
- Mass of Venus ([tex]\( m_1 \)[/tex]) = [tex]\( 4.87 \times 10^{24} \)[/tex] kg
- Mass of the probe ([tex]\( m_2 \)[/tex]) = 655 kg
- Gravitational constant ([tex]\( G \)[/tex]) = [tex]\( 6.67 \times 10^{-11} \)[/tex] N [tex]\( m^2 / kg^2 \)[/tex]
We need to find the distance [tex]\( r \)[/tex] between the probe and the center of Venus.
The formula for gravitational force is:
[tex]\[ F = G \frac{m_1 m_2}{r^2} \][/tex]
Rearranging the formula to solve for [tex]\( r \)[/tex], we get:
[tex]\[ r^2 = G \frac{m_1 m_2}{F} \][/tex]
Taking the square root of both sides, we obtain:
[tex]\[ r = \sqrt{G \frac{m_1 m_2}{F}} \][/tex]
Next, we substitute the given values into this formula:
[tex]\[ r = \sqrt{(6.67 \times 10^{-11}) \frac{(4.87 \times 10^{24}) (655)}{(2.58 \times 10^3)}} \][/tex]
Upon calculating the expression inside the square root and then taking the square root, the distance [tex]\( r \)[/tex] between the probe and the center of Venus is found to be approximately:
[tex]\[ r \approx 9,081,094.489750834 \text{ meters} \][/tex]
To convert this distance into [tex]\( 10^6 \)[/tex] meters:
[tex]\[ r_{million\_meters} = \frac{r}{10^6} \approx \frac{9,081,094.489750834}{10^6} \approx 9.081 \text{ } 10^6 \text{ meters} \][/tex]
Therefore, to three significant digits, the probe is approximately:
[tex]\[ 9.08 \times 10^6 \text{ meters} \][/tex]
from the center of Venus.
