Answer :
To determine the distance from the probe to the center of Venus, we will use the formula for gravitational force.
Given:
- Gravitational force (F): [tex]\(2.58 \times 10^3 \, \text{N}\)[/tex]
- Mass of Venus ([tex]\(m_1\)[/tex]): [tex]\(4.87 \times 10^{24} \, \text{kg}\)[/tex]
- Mass of the probe ([tex]\(m_2\)[/tex]): [tex]\(655 \, \text{kg}\)[/tex]
- Gravitational constant (G): [tex]\(6.67 \times 10^{-11} \, \text{N} \, \text{m}^2/\text{kg}^2\)[/tex]
The gravitational force formula is:
[tex]\[ F = G \frac{m_1 \cdot m_2}{r^2} \][/tex]
We need to solve for [tex]\( r \)[/tex], the distance between the probe and the center of Venus:
[tex]\[ r^2 = G \frac{m_1 \cdot m_2}{F} \][/tex]
Substituting the given values:
[tex]\[ r^2 = (6.67 \times 10^{-11} \, \text{N} \, \text{m}^2/\text{kg}^2) \frac{(4.87 \times 10^{24} \, \text{kg}) \cdot (655 \, \text{kg})}{2.58 \times 10^3 \, \text{N}} \][/tex]
[tex]\[ r^2 = 82466277131782.95 \, \text{m}^2 \][/tex]
Taking the square root of both sides to solve for [tex]\( r \)[/tex]:
[tex]\[ r = \sqrt{82466277131782.95 \, \text{m}^2} \approx 9081094.489750834 \, \text{m} \][/tex]
To express [tex]\( r \)[/tex] in terms of [tex]\( 10^6 \)[/tex] meters:
[tex]\[ r = \frac{9081094.489750834 \, \text{m}}{10^6} \approx 9.081 \times 10^6 \, \text{m} \][/tex]
Therefore, to three significant digits, the distance from the probe to the center of Venus is [tex]\( 9.081 \times 10^6 \, \text{m} \)[/tex].
Given:
- Gravitational force (F): [tex]\(2.58 \times 10^3 \, \text{N}\)[/tex]
- Mass of Venus ([tex]\(m_1\)[/tex]): [tex]\(4.87 \times 10^{24} \, \text{kg}\)[/tex]
- Mass of the probe ([tex]\(m_2\)[/tex]): [tex]\(655 \, \text{kg}\)[/tex]
- Gravitational constant (G): [tex]\(6.67 \times 10^{-11} \, \text{N} \, \text{m}^2/\text{kg}^2\)[/tex]
The gravitational force formula is:
[tex]\[ F = G \frac{m_1 \cdot m_2}{r^2} \][/tex]
We need to solve for [tex]\( r \)[/tex], the distance between the probe and the center of Venus:
[tex]\[ r^2 = G \frac{m_1 \cdot m_2}{F} \][/tex]
Substituting the given values:
[tex]\[ r^2 = (6.67 \times 10^{-11} \, \text{N} \, \text{m}^2/\text{kg}^2) \frac{(4.87 \times 10^{24} \, \text{kg}) \cdot (655 \, \text{kg})}{2.58 \times 10^3 \, \text{N}} \][/tex]
[tex]\[ r^2 = 82466277131782.95 \, \text{m}^2 \][/tex]
Taking the square root of both sides to solve for [tex]\( r \)[/tex]:
[tex]\[ r = \sqrt{82466277131782.95 \, \text{m}^2} \approx 9081094.489750834 \, \text{m} \][/tex]
To express [tex]\( r \)[/tex] in terms of [tex]\( 10^6 \)[/tex] meters:
[tex]\[ r = \frac{9081094.489750834 \, \text{m}}{10^6} \approx 9.081 \times 10^6 \, \text{m} \][/tex]
Therefore, to three significant digits, the distance from the probe to the center of Venus is [tex]\( 9.081 \times 10^6 \, \text{m} \)[/tex].
