Question

home work
\if the mass of an object increases, the gravitational force will...increase......because..............\
and \if the distance between objects increases, the gravitational force will...decrease......because..............\

Explanation:

To solve this, we use the law of universal gravitation, \( F = G\frac{m_1m_2}{r^2} \), where \( F \) is gravitational force, \( m_1, m_2 \) are masses, \( r \) is distance, and \( G \) is the gravitational constant.

For the first blank (mass increase):

Step 1: Analyze mass - force relation

From \( F = G\frac{m_1m_2}{r^2} \), if the mass (\( m_1 \) or \( m_2 \)) of an object increases (while \( r \) and the other mass are constant), the numerator \( m_1m_2 \) increases. Since \( G \) and \( r^2 \) are constant, \( F \) (gravitational force) will increase. The reason is that gravitational force is directly proportional to the product of the masses (\( F \propto m_1m_2 \) when \( r \) is constant).

For the second blank (distance increase):

Step 2: Analyze distance - force relation

From \( F = G\frac{m_1m_2}{r^2} \), if the distance (\( r \)) between objects increases (while \( m_1, m_2 \) and \( G \) are constant), the denominator \( r^2 \) increases. Since \( G, m_1, m_2 \) are constant, \( F \) (gravitational force) will decrease. The reason is that gravitational force is inversely proportional to the square of the distance (\( F \propto \frac{1}{r^2} \) when \( m_1, m_2 \) are constant).

So, the first sentence: "If the mass of an object increases, the gravitational force will increase because gravitational force is directly proportional to the product of the masses of the objects (from the law of universal gravitation \( F = G\frac{m_1m_2}{r^2} \), increasing mass increases the numerator, thus increasing force when distance is constant)."

The second sentence: "If the distance between objects increases, the gravitational force will decrease because gravitational force is inversely proportional to the square of the distance between the objects (from the law of universal gravitation \( F = G\frac{m_1m_2}{r^2} \), increasing distance increases the denominator, thus decreasing force when masses are constant)."

Answer:

To solve this, we use the law of universal gravitation, \( F = G\frac{m_1m_2}{r^2} \), where \( F \) is gravitational force, \( m_1, m_2 \) are masses, \( r \) is distance, and \( G \) is the gravitational constant.

For the first blank (mass increase):

Step 1: Analyze mass - force relation

From \( F = G\frac{m_1m_2}{r^2} \), if the mass (\( m_1 \) or \( m_2 \)) of an object increases (while \( r \) and the other mass are constant), the numerator \( m_1m_2 \) increases. Since \( G \) and \( r^2 \) are constant, \( F \) (gravitational force) will increase. The reason is that gravitational force is directly proportional to the product of the masses (\( F \propto m_1m_2 \) when \( r \) is constant).

For the second blank (distance increase):

Step 2: Analyze distance - force relation

From \( F = G\frac{m_1m_2}{r^2} \), if the distance (\( r \)) between objects increases (while \( m_1, m_2 \) and \( G \) are constant), the denominator \( r^2 \) increases. Since \( G, m_1, m_2 \) are constant, \( F \) (gravitational force) will decrease. The reason is that gravitational force is inversely proportional to the square of the distance (\( F \propto \frac{1}{r^2} \) when \( m_1, m_2 \) are constant).

So, the first sentence: "If the mass of an object increases, the gravitational force will increase because gravitational force is directly proportional to the product of the masses of the objects (from the law of universal gravitation \( F = G\frac{m_1m_2}{r^2} \), increasing mass increases the numerator, thus increasing force when distance is constant)."

The second sentence: "If the distance between objects increases, the gravitational force will decrease because gravitational force is inversely proportional to the square of the distance between the objects (from the law of universal gravitation \( F = G\frac{m_1m_2}{r^2} \), increasing distance increases the denominator, thus decreasing force when masses are constant)."