Looking for trustworthy answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Explore thousands of questions and answers from a knowledgeable community of experts on our user-friendly platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our 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. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.
