Problem 62 - Master Resource Book in JEE Main Physics

Question

Two bodies of masses \(2 \mathrm{~kg}\) and \(8 \mathrm{~kg}\) are separated by a distance of \(9 \mathrm{~m}\). The point where the resultant gravitational field intensity is zero is at a distance of (a) \(4.5 \mathrm{~m}\) from each mass (b) \(6 \mathrm{~m}\) from \(2 \mathrm{~kg}\) (c) \(6 \mathrm{~m}\) from \(8 \mathrm{~kg}\) (d) \(2.5 \mathrm{~m}\) from \(2 \mathrm{~kg}\)

Step-by-Step Solution

Answer

The correct distance is (b) 6 m from 2 kg.

Step 1: Identify the concept

This problem involves finding the point between two masses where the gravitational field intensity is zero. This occurs where the gravitational forces exerted by the two masses cancel each other out.

Step 2: Apply the gravitational force equation

The gravitational force exerted by a mass is given by the formula: \[ F = \frac{G imes m}{r^2} \]where \(F\) is the force, \(G\) is the gravitational constant, \(m\) is the mass, and \(r\) is the distance from the mass.

Step 3: Set up the equation for forces to be equal

For the point where the gravitational field intensity is zero, the gravitational forces due to the two masses must be equal in magnitude:\[ \frac{G \times 2}{x^2} = \frac{G \times 8}{(9-x)^2} \]where \(x\) is the distance from the 2 kg mass.

Step 4: Simplify the equation

Since \(G\) is a common factor, it can be canceled out from both sides of the equation:\[ \frac{2}{x^2} = \frac{8}{(9-x)^2} \]

Step 5: Cross multiply and solve for x

Cross multiply to remove the fractions:\[ 2(9-x)^2 = 8x^2 \]Simplifying further:\[ 2(81 - 18x + x^2) = 8x^2 \]\[ 162 - 36x + 2x^2 = 8x^2 \]

Step 6: Rearrange the equation

Bring all terms to one side to form a quadratic equation:\[ 2x^2 - 8x^2 - 36x + 162 = 0 \]\[ -6x^2 - 36x + 162 = 0 \]Divide the equation by -2:\[ 3x^2 + 18x - 81 = 0 \]

Step 7: Solve the quadratic equation

Use the quadratic formula where \(a = 3\), \(b = 18\), \(c = -81\):\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]\[ x = \frac{-18 \pm \sqrt{18^2 - 4 \times 3 \times -81}}{2 \times 3} \]Simplify under the square root and solve for \(x\).

Step 8: Check the solutions

From calculating the quadratic formula, we obtain possible \(x\) values. Verify which value fits the situation by checking it divides the total distance of 9 m between the two mass points accurately based on context.

Key Concepts

Gravitational Force

Gravitational force is a fundamental force of nature. It is the attractive force exerted by objects with mass. This force pulls objects towards each other.
You may remember the gravitational force formula:

• The force (\( F \)) between two masses, \
• The masses (\( m_1 \) and \( m_2 \)), and
• The distance (\( r \)) separating them.

Working together, these form the equation:\( F = \frac{G \times m_1 \times m_2}{r^2} \), where \( G \)is the gravitational constant. This formula tells us that gravitational force is stronger when objects are closer together or have greater mass. As we explore the specific problem between two particular masses, we apply this law to find a point where their gravitational attractions balance out.

Point of Zero Gravitational Intensity

The point of zero gravitational intensity is an intriguing concept. It's the point between two masses where gravitational forces cancel each other out. In simpler terms, it's like standing in a place where you don't feel any pull from nearby objects.
When dealing with two masses, say \( m_1 \) and \( m_2 \), there exists such a point along the line connecting them. At this point, the attractive force from \( m_1 \) is exactly equal and opposite to that from \( m_2 \). This ensures that the net force you would experience is zero.
In our particular exercise, finding such a point involves setting up an equation based on the gravitational forces each mass exerts, which is then solved to identify how far this point is from the known objects.

Quadratic Equation

A quadratic equation is a type of polynomial equation that is critical in solving many physical problems, including our current one. Its general form is \( ax^2 + bx + c = 0 \).The objective is to find values of \( x \) that satisfy the equation.
These equations often appear when forces or distances involve square terms, common in gravitational problems like the one we're discussing. We learned that after simplifying our force equation by eliminating common terms, we derived such a quadratic equation:\[ 3x^2 + 18x - 81 = 0 \]. Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), we can solve for \( x \). This formula lets us find potential points where the gravitational forces from two masses balance each other.

Mass and Distance Relationship

The relationship between mass and distance is crucial in gravitational interactions. It's embedded in the formula \( F = \frac{G \times m}{r^2} \), showing how force changes with distance.
Two important concepts arise from this:

• Increasing mass increases gravitational force.
• Increasing distance decreases gravitational force.

When evaluating multiple masses, like in our specific case, understanding the interplay of mass and distance determines how each mass's gravity influences the space around it. Finding a point of zero gravitational intensity is essentially finding a spot where their distances inversely balance out their gravitational pulls.
As we set up equations for our exercise, this relationship was key in establishing where gravitational equilibrium, or zero intensity, occurs, guiding us through solving the problem efficiently.