Question

A mass \(m\) is in a uniform gravitational field, which exerts the usual force \(F=m g\) vertically down, but with \(g\) varying with time, \(g=g(t) .\) Choosing axes with \(y\) measured vertically up and defining \(U=m g y\) as usual, show that \(\mathbf{F}=-\nabla U\) as usual, but, by differentiating \(E=\frac{1}{2} m v^{2}+U\) with respect to \(t,\) show that \(E\) is not conserved.

Short answer

Energy is not conserved if \( g(t) \) changes over time, as shown by \( \frac{dE}{dt} = m\frac{dg}{dt}y \neq 0 \).

Step-by-step solution

Define Potential Energy

The potential energy in a uniform gravitational field where the gravitational acceleration \( g \) varies with time \( t \) is given by \( U = mgy \). The forces are defined on the axes such that the \( y \) axis points vertically upwards.

Calculate the Gradient of Potential Energy

The gradient \( abla U \) in this context where we only consider the vertical direction is the partial derivative with respect to \( y \). Therefore, \[ abla U = \frac{\partial U}{\partial y} = \frac{\partial}{\partial y} (mgy) = mg. \]

Set Up Force Equation

The force \( \mathbf{F} \) in the gravitational field is \( F = mg \). According to the equation \( \mathbf{F} = -abla U \), we substitute the gradient calculated: \[ F = -mg. \] This shows that the standard force expression \( F = -abla U \) holds true.

Differentiate the Energy Expression

Consider the total energy \( E = \frac{1}{2} mv^{2} + U \). Differentiating this expression with respect to time \( t \), we get: \[ \frac{dE}{dt} = \frac{d}{dt} \left( \frac{1}{2} mv^{2} \right) + \frac{d}{dt} (mgy). \]

Differentiate the Kinetic Energy

Differentiating the kinetic energy part gives: \[ \frac{d}{dt} \left( \frac{1}{2} mv^{2} \right) = m v \frac{dv}{dt} = mv a, \] where \( a \) is the acceleration, which is \( -g(t) \) as gravity is acting downward.

Differentiate the Potential Energy

Differentiating the potential energy part: \[ \frac{d}{dt} (mgy) = m \left( \frac{dg}{dt} y + g \frac{dy}{dt} \right) = m \frac{dg}{dt} y + mgv. \]

Combine Energy Derivatives

Combine the differentiated kinetic and potential energy expressions: \[ \frac{dE}{dt} = mva - mgv + m \frac{dg}{dt} y + mgv. \] The \( mgv \) terms cancel each other out.

Simplify and Conclude Non-conservation

The resulting expression is \[ \frac{dE}{dt} = mva + m \frac{dg}{dt} y. \] Since \( ma = -mg \), \( \frac{dE}{dt} = 0 + m \frac{dg}{dt} y = m \frac{dg}{dt} y \), showing \( \frac{dE}{dt} eq 0 \) unless \( \frac{dg}{dt} = 0 \), the energy is not conserved if \( g(t) \) is changing with time.

Key concepts

Potential Energy

Potential energy is the stored energy of an object due to its position in a gravitational field. In this scenario, potential energy is given by the formula \( U = mgy \), where \( m \) is the mass of the object, \( g \) is the gravitational field strength, and \( y \) is the height above the reference point (usually ground level).
This formula assumes that the gravitational force acts vertically downward. Thus, we define the positive direction for potential energy as vertically upward.
An interesting aspect of this problem is that \( g \), the gravitational field strength, varies with time. This means the potential energy changes as time progresses, even if the position, \( y \), remains constant. Since potential energy depends not only on position but also on the strength of the gravitational field, the time-varying \( g \) directly impacts the potential energy value.

Gravitational Field

A gravitational field is a region of space around a mass where another mass experiences a force due to gravity.
Usually, we consider gravitational field strength, denoted by \( g \), to be a constant value. However, in this exercise, the gravitational field strength changes over time, denoted as \( g(t) \).
The force exerted by a gravitational field on a mass is calculated using the formula \( F = mg \), meaning the force depends on both the mass and the field strength. In our situation, this force remains directed downward, parallel to the traditional setup where the gravitational field is uniform but constant.

Energy Conservation

Energy conservation is a fundamental principle stating that the total energy in an isolated system remains constant over time. Normally, we would expect the total mechanical energy, the sum of kinetic and potential energy, to be conserved.
In our problem, the total energy \( E \) is given by \( E = \frac{1}{2} mv^2 + U \), which includes both the kinetic energy and the potential energy.
Differentiating this expression with respect to time reveals that \( \frac{dE}{dt} \) is not zero unless \( \frac{dg}{dt} = 0 \). This insight tells us that energy is not conserved if \( g \) varies with time.

• When \( g \) is constant over time, potential and kinetic energies change in a way that keeps total energy the same.
• If \( g \)'s value changes, our system gains or loses energy in response to how \( g(t) \) changes, thus leading to energy non-conservation.

This elucidates how the variability of external conditions such as the gravitational field strength can impact energy conservation principles.