Westonci.ca offers quick and accurate answers to your questions. Join our community and get the insights you need today. Experience the convenience of finding accurate answers to your questions from knowledgeable professionals on our platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly 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.
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.
