Question

Two identical 20.0 -kg spheres of radius \(10 \mathrm{~cm}\) are \(30.0 \mathrm{~cm}\) apart (center-to-center distance). a) If they are released from rest and allowed to fall toward one another, what is their speed when they first make contact? b) If the spheres are initially at rest and just touching, how much energy is required to separate them to \(1.00 \mathrm{~m}\) apart? Assume that the only force acting on each mass is the gravitational force due to the other mass.

Short answer

Answer: The speed when the spheres first make contact is approximately \(0.036 \mathrm{m/s}\) and the energy required to separate the spheres to \(1.00 \mathrm{~m}\) apart is \(3.34 \times 10^{-9} \mathrm{~J}\).

Step-by-step solution

Identify the relevant values given in the problem.

In this case, we have the masses of both spheres (\(m_1=m_2=20.0 \mathrm{~kg}\)), radius of the spheres (\(r=10 \mathrm{~cm}=0.1 \mathrm{~m}\)), and the initial distance between the centers of the spheres (\(d_{initial}=30.0 \mathrm{~cm}=0.3 \mathrm{~m}\)).

Write the gravitational force formula and expression for potential energy.

We know that the gravitational force between two masses is given by \(F=\displaystyle\frac{G\cdot m_1 \cdot m_2}{d^2}\), where \(G = 6.674 \times 10^{-11} \mathrm{m^3 \cdot kg^{-1} \cdot s^{-2}}\) is the gravitational constant. The expression for potential energy due to gravity is \(U = \displaystyle\frac{G\cdot m_1 \cdot m_2 }{d}\).

Use the conservation of energy to find the speed when the spheres make contact.

The Law of Conservation of Energy states that the total mechanical energy in a system remains constant in the absence of external forces. In this case, only gravitational forces are acting, so the conservation of energy is given as: \(\frac{1}{2} m (\upsilon_f^2 - \upsilon_i^2) = -\Delta U\) Initially, the spheres are at rest (\(\upsilon_i = 0\)), and the final distance between the centers of the spheres when they just make contact is given by \(d_{final} = 2r=0.2\mathrm{~m}\). Using the expression for potential energy, we have \(\Delta U = U_{final}-U_{initial}= -G\cdot m_1 \cdot m_2 \cdot \displaystyle\left(\frac{1}{d_{final}} - \frac{1}{d_{initial}}\right)\). Now the conservation of energy equation can be written as: \(\frac{1}{2} m \upsilon_f^2 = G\cdot m_1 \cdot m_2 \cdot \displaystyle\left(\frac{1}{d_{final}} - \frac{1}{d_{initial}}\right)\).

Compute the speed when the spheres first make contact.

We will now solve the equation we derived in step 3 for \(\upsilon_f\): \(\upsilon_f = \sqrt{\displaystyle\frac{2G\cdot m_1 \cdot m_2 \cdot \left(\frac{1}{d_{final}} - \frac{1}{d_{initial}}\right)}{m}}\). Plug in the given values and compute for \(\upsilon_f\): \(\upsilon_f = \sqrt{\displaystyle\frac{2(6.674\times 10^{-11} \mathrm{m^3 \cdot kg^{-1} \cdot s^{-2}})(20.0\mathrm{~kg})^2(\frac{1}{0.2\mathrm{~m}} - \frac{1}{0.3\mathrm{~m}})}{20.0\mathrm{~kg}}}\). Solving for \(\upsilon_f ≈ 0.036 \mathrm{m/s}\). #b)#

Use the same potential energy equation for separation.

To separate the spheres to \(1.00 \mathrm{~m}\) apart, we need to change their potential energy. We can find the required energy by finding the change in potential energy between the spheres just touching and being \(1.00 \mathrm{~m}\) apart.

Calculate the difference in potential energy.

Using the same expression for the potential energy difference as in part a, we have \(\Delta U = -G\cdot m_1 \cdot m_2 \cdot \displaystyle\left(\frac{1}{1.00 \mathrm{~m}} - \frac{1}{0.2\mathrm{~m}}\right)\).

Compute the energy required to separate the spheres.

Plug in the given values and compute for the energy required: \(\Delta U = -(6.674\times 10^{-11} \mathrm{m^3 \cdot kg^{-1} \cdot s^{-2}})(20.0\mathrm{~kg})^2(\frac{1}{1.00 \mathrm{~m}} - \frac{1}{0.2\mathrm{~m}}) ≈ 3.34 \times 10^{-9} \mathrm{~J}\). So, the energy required to separate the spheres to \(1.00 \mathrm{~m}\) apart is \(3.34 \times 10^{-9} \mathrm{~J}\).

Key concepts

Gravitational Force

The concept of gravitational force is pivotal to understanding the movement of objects in space and their interaction with each other. It’s this invisible force that keeps our feet firmly on the ground and governs the motion of planets in their orbits. A key factor in gravitational force is mass; the greater the mass of an object, the stronger its gravitational pull.

Newton's law of universal gravitation, which will be discussed later, first introduced the calculation of this force. In essence, the equation implies that every mass exerts an attractive force on every other mass. This force is more pronounced when objects are closer together. For instance, in the provided exercise, two spheres attract each other due to the gravitational force between them, despite their relatively small size and mass compared to celestial bodies like planets and stars.

Understanding the gravitational force helps in predicting how the spheres in the problem will move towards each other once released. The strength of this force decreases rapidly with distance, which influences not just the motion but also the potential energy between the objects, setting the stage for concepts such as gravitational potential energy and conservation of energy.

Conservation of Energy

The law of conservation of energy is a fundamental principle in physics that states that energy cannot be created or destroyed, only transferred or transformed from one form to another. In the context of our textbook problem, we utilize this principle by looking at how the gravitational potential energy of the two spheres is converted into kinetic energy as they move toward each other.

Initially, when the spheres are at rest and separated, they possess potential energy due to their position in a gravitational field. As they fall toward one another, this potential energy is converted to kinetic energy, which is the energy of motion. By the time they touch, all of the initial potential energy assumes the form of kinetic energy (assuming there are no other forces at work such as friction or air resistance).

The conservation of energy is central to solving part (a) of our exercise, where we derive the speed of the spheres just before contact by equating the decrease in potential energy to the increase in kinetic energy. This approach is key to understanding a wide array of physical phenomena beyond merely answering a homework question.

Newton's Law of Gravitation

Sir Isaac Newton revolutionized our understanding of the universe with his law of universal gravitation. This law states that every point mass attracts every other point mass in the universe with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers. Mathematically, it's expressed as:
\[ F = G\frac{m_1 \cdot m_2}{d^2} \]
Here, \( F \) is the gravitational force, \( G \) is the gravitational constant \( (6.674 \times 10^{-11} \, \text{m}^3 \cdot \text{kg}^{-1} \cdot \text{s}^{-2}) \), \( m_1 \) and \( m_2 \) are the masses of the two objects, and \( d \) is the distance between their centers.

In our exercise, Newton's law of gravitation allows us to calculate the gravitational force between the two spheres and, with that, the gravitational potential energy. Although we are dealing with spheres of modest mass, the same law governs the dynamics of galaxies, illustrating the universal applicability of Newton's groundbreaking insight.

Kinetic Energy

Kinetic energy is the energy that an object possesses due to its motion, and it plays a central role in the narrative of physics. The amount of kinetic energy an object has is dependent on its mass and the square of its velocity, following the formula:
\[ KE = \frac{1}{2} mv^2 \]
In this formula, \( KE \) represents kinetic energy, \( m \) is the mass of the object, and \( v \) is its velocity.

Relating this to our exercise, as the gravitational potential energy of the two spheres decreases, their kinetic energy increases due to their acceleration toward each other under the influence of gravity. By the time they touch, assuming no energy losses, the kinetic energy will equal the initial potential energy they had when they were apart. This conversion is beautifully exemplified when calculating the final speed of the spheres just before contact, underscoring the transformation from potential to kinetic energy as guided by the conservation of energy.