Question

A hammer with mass \(m\) is dropped from rest from a height \(h\) above the earth's surface. This height is not necessarily small compared with the radius \(R_E\) of the earth. Ignoring air resistance, derive an expression for the speed y of the hammer when it reaches the earth's surface. Your expression should involve \(h\), \(R_E\), and \(m_E\) (the earth's mass).

Short answer

The speed of the hammer is \( v = \sqrt{\frac{2G m_E h}{R_E (R_E + h)}} \)."

Step-by-step solution

Gravitational Potential Energy at Height

The gravitational potential energy of the hammer at height \(h\) is given by the expression: \[ U_i = -\frac{G m m_E}{R_E + h} \] where \(G\) is the gravitational constant.

Gravitational Potential Energy at Surface

As the hammer reaches the Earth's surface, at \(R_E\), the gravitational potential energy is: \[ U_f = -\frac{G m m_E}{R_E} \]

Conservation of Mechanical Energy

According to the conservation of mechanical energy, the initial potential energy plus initial kinetic energy equals the final total energy. Since it starts from rest, we have: \[ U_i + K_i = U_f + K_f \] which simplifies to \[ -\frac{G m m_E}{R_E + h} + 0 = -\frac{G m m_E}{R_E} + \frac{1}{2} m v^2 \]

Solve for Kinetic Energy

Rearranging the conservation of energy equation gives us: \[ \frac{1}{2} mv^2 = \left(-\frac{G m m_E}{R_E} + \frac{G m m_E}{R_E + h}\right) \]

Simplify and Solve for Velocity

Simplify the equation: \[ \frac{1}{2} mv^2 = \frac{G m m_E}{R_E + h} - \frac{G m m_E}{R_E} \] \[ v^2 = 2G m_E \left(\frac{1}{R_E} - \frac{1}{R_E + h}\right) \] \[ v^2 = 2G m_E \left(\frac{R_E + h - R_E}{R_E (R_E + h)}\right) \] \[ v^2 = 2G m_E \left(\frac{h}{R_E (R_E + h)}\right) \]Thus, solving for \(v\), we have: \[ v = \sqrt{\frac{2G m_E h}{R_E (R_E + h)}} \]

Key concepts

Conservation of Mechanical Energy

One of the core ideas in physics is the conservation of mechanical energy. This principle posits that the total mechanical energy (the sum of potential and kinetic energy) in an isolated system remains constant if only conservative forces, such as gravity, act on it.
When the hammer is dropped from a height, its initial energy is purely gravitational potential energy because it starts from rest, meaning it has no kinetic energy initially. As the hammer falls, it loses potential energy, which transfers to kinetic energy as the hammer speeds up. By the time the hammer reaches the Earth's surface, all the gravitational potential energy has been converted into kinetic energy.
This concept can be represented mathematically through the equation:

• Initial Potential Energy + Initial Kinetic Energy = Final Potential Energy + Final Kinetic Energy

Since the hammer is dropped from rest, the initial kinetic energy is zero, simplifying our calculations significantly.

Gravitational Constant

The gravitational constant, denoted as \(G\), is a key factor in the calculations involving gravitational forces between two masses. It represents the strength of gravity on a universal scale. The gravitation constant has a value of approximately \(6.674 \times 10^{-11} \text{Nm}^2/\text{kg}^2\), which indicates the small magnitude of the gravitational force compared to other fundamental forces such as electromagnetism.
In this exercise, \(G\) is used to determine the gravitational potential energy of the hammer both at its initial height and at the Earth's surface. This involves the formula:

• Gravitational Potential Energy at a distance \(r\) from the Earth's center: \(-\frac{G m m_E}{r}\)

The gravitational constant is crucial for calculating the gravitational interaction in various contexts, spanning from small-scale physical objects to celestial bodies.

Kinetic Energy

Kinetic energy is the energy that an object possesses due to its motion. When the hammer is released and falls under the force of gravity, it accelerates towards the Earth, gaining velocity. This increase in velocity translates the potential energy it had while at rest at a height into kinetic energy.
The formula for kinetic energy is given by:

• \( K = \frac{1}{2} mv^2 \)

where \(m\) is the mass of the hammer and \(v\) is its velocity.
As the hammer falls, the potential energy decreases and its kinetic energy increases, conserving the total mechanical energy of the system. This balance continues until the hammer makes contact with the Earth's surface.

Earth's Radius

The radius of Earth, denoted as \(R_E\), is an essential parameter when calculating gravitational potential energy for objects located at a significant height from the planet’s surface. It measures approximately 6,371 kilometers and plays a vital role in determining the gravitational effects experienced by objects within the Earth's gravitational field.
In our exercise, Earth's radius is used to calculate the initial and final gravitational potential energies of the hammer. The initial potential energy accounts for the distance of \(R_E + h\), while the final potential energy considers the distance \(R_E\).
Understanding and incorporating Earth’s radius is critical, especially when dealing with space-related physics problems, as it outlines the scale and influence of gravitational forces acting at various distances from the Earth's center.