Welcome to Westonci.ca, the ultimate question and answer platform. Get expert answers to your questions quickly and accurately. Join our Q&A platform and get accurate answers to all your questions from professionals across multiple disciplines. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
To determine how many times greater the gravitational force between the Sun and Mercury is compared to the force between the Sun and Pluto, we need to apply Newton's Law of Gravitation. According to this law, the gravitational force [tex]\( F \)[/tex] between two objects can be represented as:
[tex]\[ F = G \frac{m_1 m_2}{r^2} \][/tex]
where:
- [tex]\( G \)[/tex] is the universal gravitational constant,
- [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] are the masses of the two objects,
- [tex]\( r \)[/tex] is the distance between the centers of the two objects.
Let's apply this to our specific situation. Let:
- [tex]\( m_{\text{mercury}} \)[/tex] be the mass of Mercury,
- [tex]\( m_{\text{pluto}} \)[/tex] be the mass of Pluto,
- [tex]\( d_{\text{mercury}} \)[/tex] be the distance from the Sun to Mercury,
- [tex]\( d_{\text{pluto}} \)[/tex] be the distance from the Sun to Pluto.
Given that:
- The mass of Mercury is 2.2 times the mass of Pluto:
[tex]\[ m_{\text{mercury}} = 2.2 \times m_{\text{pluto}} \][/tex]
- Pluto is 102.1 times as far from the Sun as Mercury:
[tex]\[ d_{\text{pluto}} = 102.1 \times d_{\text{mercury}} \][/tex]
We want to find the ratio of the gravitational force between the Sun and Mercury ([tex]\( F_{\text{mercury}} \)[/tex]) to the gravitational force between the Sun and Pluto ([tex]\( F_{\text{pluto}} \)[/tex]). Using Newton's law of gravitation for each case, we have:
[tex]\[ F_{\text{mercury}} = G \frac{m_{\text{sun}} m_{\text{mercury}}}{d_{\text{mercury}}^2} \][/tex]
[tex]\[ F_{\text{pluto}} = G \frac{m_{\text{sun}} m_{\text{pluto}}}{d_{\text{pluto}}^2} \][/tex]
To find the force ratio, we divide these two equations:
[tex]\[ \frac{F_{\text{mercury}}}{F_{\text{pluto}}} = \frac{G \frac{m_{\text{sun}} m_{\text{mercury}}}{d_{\text{mercury}}^2}}{G \frac{m_{\text{sun}} m_{\text{pluto}}}{d_{\text{pluto}}^2}} \][/tex]
Since [tex]\( G \)[/tex] and [tex]\( m_{\text{sun}} \)[/tex] are constants and appear in both the numerator and the denominator, they cancel out:
[tex]\[ \frac{F_{\text{mercury}}}{F_{\text{pluto}}} = \frac{m_{\text{mercury}} / d_{\text{mercury}}^2}{m_{\text{pluto}} / d_{\text{pluto}}^2} \][/tex]
Substitute the given ratios for [tex]\( m_{\text{mercury}} \)[/tex] and [tex]\( d_{\text{pluto}} \)[/tex]:
[tex]\[ \frac{F_{\text{mercury}}}{F_{\text{pluto}}} = \frac{(2.2 \times m_{\text{pluto}}) / d_{\text{mercury}}^2}{m_{\text{pluto}} / (102.1 \times d_{\text{mercury}})^2} \][/tex]
Simplify the expression:
[tex]\[ \frac{F_{\text{mercury}}}{F_{\text{pluto}}} = \frac{2.2 \times m_{\text{pluto}} / d_{\text{mercury}}^2}{m_{\text{pluto}} / (102.1^2 \times d_{\text{mercury}}^2)} \][/tex]
The [tex]\( m_{\text{pluto}} \)[/tex] in the numerator and the denominator cancel out, and [tex]\( d_{\text{mercury}}^2 \)[/tex] in the numerator and the denominator also cancel out:
[tex]\[ \frac{F_{\text{mercury}}}{F_{\text{pluto}}} = \frac{2.2}{102.1^2} \][/tex]
You would then calculate:
[tex]\[ \frac{2.2}{102.1^2} = 0.00021104311898706982 \][/tex]
Thus, the gravitational force between the Sun and Mercury is approximately [tex]\( 0.000211 \)[/tex] times that of the gravitational force between the Sun and Pluto.
[tex]\[ F = G \frac{m_1 m_2}{r^2} \][/tex]
where:
- [tex]\( G \)[/tex] is the universal gravitational constant,
- [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] are the masses of the two objects,
- [tex]\( r \)[/tex] is the distance between the centers of the two objects.
Let's apply this to our specific situation. Let:
- [tex]\( m_{\text{mercury}} \)[/tex] be the mass of Mercury,
- [tex]\( m_{\text{pluto}} \)[/tex] be the mass of Pluto,
- [tex]\( d_{\text{mercury}} \)[/tex] be the distance from the Sun to Mercury,
- [tex]\( d_{\text{pluto}} \)[/tex] be the distance from the Sun to Pluto.
Given that:
- The mass of Mercury is 2.2 times the mass of Pluto:
[tex]\[ m_{\text{mercury}} = 2.2 \times m_{\text{pluto}} \][/tex]
- Pluto is 102.1 times as far from the Sun as Mercury:
[tex]\[ d_{\text{pluto}} = 102.1 \times d_{\text{mercury}} \][/tex]
We want to find the ratio of the gravitational force between the Sun and Mercury ([tex]\( F_{\text{mercury}} \)[/tex]) to the gravitational force between the Sun and Pluto ([tex]\( F_{\text{pluto}} \)[/tex]). Using Newton's law of gravitation for each case, we have:
[tex]\[ F_{\text{mercury}} = G \frac{m_{\text{sun}} m_{\text{mercury}}}{d_{\text{mercury}}^2} \][/tex]
[tex]\[ F_{\text{pluto}} = G \frac{m_{\text{sun}} m_{\text{pluto}}}{d_{\text{pluto}}^2} \][/tex]
To find the force ratio, we divide these two equations:
[tex]\[ \frac{F_{\text{mercury}}}{F_{\text{pluto}}} = \frac{G \frac{m_{\text{sun}} m_{\text{mercury}}}{d_{\text{mercury}}^2}}{G \frac{m_{\text{sun}} m_{\text{pluto}}}{d_{\text{pluto}}^2}} \][/tex]
Since [tex]\( G \)[/tex] and [tex]\( m_{\text{sun}} \)[/tex] are constants and appear in both the numerator and the denominator, they cancel out:
[tex]\[ \frac{F_{\text{mercury}}}{F_{\text{pluto}}} = \frac{m_{\text{mercury}} / d_{\text{mercury}}^2}{m_{\text{pluto}} / d_{\text{pluto}}^2} \][/tex]
Substitute the given ratios for [tex]\( m_{\text{mercury}} \)[/tex] and [tex]\( d_{\text{pluto}} \)[/tex]:
[tex]\[ \frac{F_{\text{mercury}}}{F_{\text{pluto}}} = \frac{(2.2 \times m_{\text{pluto}}) / d_{\text{mercury}}^2}{m_{\text{pluto}} / (102.1 \times d_{\text{mercury}})^2} \][/tex]
Simplify the expression:
[tex]\[ \frac{F_{\text{mercury}}}{F_{\text{pluto}}} = \frac{2.2 \times m_{\text{pluto}} / d_{\text{mercury}}^2}{m_{\text{pluto}} / (102.1^2 \times d_{\text{mercury}}^2)} \][/tex]
The [tex]\( m_{\text{pluto}} \)[/tex] in the numerator and the denominator cancel out, and [tex]\( d_{\text{mercury}}^2 \)[/tex] in the numerator and the denominator also cancel out:
[tex]\[ \frac{F_{\text{mercury}}}{F_{\text{pluto}}} = \frac{2.2}{102.1^2} \][/tex]
You would then calculate:
[tex]\[ \frac{2.2}{102.1^2} = 0.00021104311898706982 \][/tex]
Thus, the gravitational force between the Sun and Mercury is approximately [tex]\( 0.000211 \)[/tex] times that of the gravitational force between the Sun and Pluto.
Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.
