Q.3 a. What is the difference between the force in F =ma (Newton's 2nd law) and F = Gm₁m₂/d² (Newton's law of universal gravitation). b. Find the net force on a Planet A of mass 3.0 x 10¹ kg due to the gravitational attraction of both Planet B of mass 4.9 x 10²0 kg and the Sun of mass 2.1 x 10³⁰ kg, assuming they are at right angles to each other. The distance of Planet A from Planet B is 5.0 x 107 m and distance of the Sun from the Planet A is 5.8 x 106 m. G= 6.67 x 10-¹¹ N.m²/kg².
Q.3 a. What is the difference between the force in F =ma (Newton's 2nd law) and F = Gm₁m₂/d² (Newton's law of universal gravitation). b. Find the net force on a Planet A of mass 3.0 x 10¹ kg due to the gravitational attraction of both Planet B of mass 4.9 x 10²0 kg and the Sun of mass 2.1 x 10³⁰ kg, assuming they are at right angles to each other. The distance of Planet A from Planet B is 5.0 x 107 m and distance of the Sun from the Planet A is 5.8 x 106 m. G= 6.67 x 10-¹¹ N.m²/kg².
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Chapter1: Units, Trigonometry. And Vectors
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![### Question 3
#### a. What is the difference between the force in \( F = ma \) (Newton's 2nd law) and \( F = \frac{Gm_1m_2}{d^2} \) (Newton’s law of universal gravitation)?
Newton's Second Law of Motion, expressed as \( F = ma \), states that the force \( F \) acting on an object is equal to the mass \( m \) of the object multiplied by its acceleration \( a \).
Newton’s Law of Universal Gravitation, expressed as \( F = \frac{Gm_1m_2}{d^2} \), states that every point mass attracts every other point mass by a force acting along the line intersecting both points. The formula calculates the gravitational force \( F \) between two masses (\( m_1 \) and \( m_2 \)) separated by a distance \( d \), with \( G \) being the gravitational constant.
#### b. Find the net force on a Planet A of mass \( 3.0 \times 10^{19} \) kg due to the gravitational attraction of both Planet B of mass \( 4.9 \times 10^{20} \) kg and the Sun of mass \( 2.1 \times 10^{30} \) kg, assuming they are at right angles to each other. The distance of Planet A from Planet B is \( 5.0 \times 10^7 \) m and the distance of the Sun from Planet A is \( 5.8 \times 10^6 \) m. \( G = 6.67 \times 10^{-11} \) N·m²/kg².
To solve for the net force:
1. **Force between Planet A and Planet B:**
\[
F_{AB} = \frac{G m_A m_B}{d_{AB}^2}
\]
where
\( m_A = 3.0 \times 10^{19} \) kg,
\( m_B = 4.9 \times 10^{20} \) kg,
\( d_{AB} = 5.0 \times 10^7 \) m.
\[
F_{AB} = \frac{6.67 \times 10^{-11} \times 3.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F64a30517-e863-4662-9804-11d4230920b7%2F10c711cf-485d-4daa-a971-ea460f3eb17d%2Fiekarm_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Question 3
#### a. What is the difference between the force in \( F = ma \) (Newton's 2nd law) and \( F = \frac{Gm_1m_2}{d^2} \) (Newton’s law of universal gravitation)?
Newton's Second Law of Motion, expressed as \( F = ma \), states that the force \( F \) acting on an object is equal to the mass \( m \) of the object multiplied by its acceleration \( a \).
Newton’s Law of Universal Gravitation, expressed as \( F = \frac{Gm_1m_2}{d^2} \), states that every point mass attracts every other point mass by a force acting along the line intersecting both points. The formula calculates the gravitational force \( F \) between two masses (\( m_1 \) and \( m_2 \)) separated by a distance \( d \), with \( G \) being the gravitational constant.
#### b. Find the net force on a Planet A of mass \( 3.0 \times 10^{19} \) kg due to the gravitational attraction of both Planet B of mass \( 4.9 \times 10^{20} \) kg and the Sun of mass \( 2.1 \times 10^{30} \) kg, assuming they are at right angles to each other. The distance of Planet A from Planet B is \( 5.0 \times 10^7 \) m and the distance of the Sun from Planet A is \( 5.8 \times 10^6 \) m. \( G = 6.67 \times 10^{-11} \) N·m²/kg².
To solve for the net force:
1. **Force between Planet A and Planet B:**
\[
F_{AB} = \frac{G m_A m_B}{d_{AB}^2}
\]
where
\( m_A = 3.0 \times 10^{19} \) kg,
\( m_B = 4.9 \times 10^{20} \) kg,
\( d_{AB} = 5.0 \times 10^7 \) m.
\[
F_{AB} = \frac{6.67 \times 10^{-11} \times 3.
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