Question

The force of gravitational attraction on a mass \(m\) at distance \(r\) from the center of the earth \((r>\text { radius } R \text { of the earth })\) is $$m g R^{2} / r^{2}$$. Then the differential equation of motion of a mass \(m\) projected radially outward from the surface of the earth, with initial velocity \(v_{0},\) is $$m d^{2} r / d t^{2}=-m g R^{2} / r^{2}$$. Use method (c) above to find \(v\) as a function of \(r\) if \(v=v_{0}\) initially (that is, when \(r=R) .\) Find the maximum value of \(r\) for a given \(v_{0},\) that is, the value of \(r\) when \(v=0 .\) Find the escape velocity, that is, the smallest value of \(v_{0}\) for which \(r\) can tend to infinity.

Short answer

The maximum value of \( r \) is \( \frac{R}{1 - \frac{v_0^2}{2gR}} \. The escape velocity is \( \sqrt{2gR} \).

Step-by-step solution

Understanding the Differential Equation

The given differential equation is \( m \frac{d^2 r}{dt^2} = -m \frac{g R^2}{r^2} \). This represents the motion of a mass projected radially outward from the surface of the Earth.

Simplifying and Eliminating Mass

Since mass \(m\) is present in every term, it can be divided out, giving: \[ \frac{d^2 r}{dt^2} = - \frac{g R^2}{r^2} \]

Using Energy Conservation

Utilize the principle of conservation of energy: the sum of kinetic and potential energy remains constant. Kinetic energy is given by \( \frac{1}{2} mv^2 \), and the gravitational potential energy at a distance \(r\) is \( - \frac{GMm}{r} \).

Setting Up Energy Equations

At \(r = R\), the initial energy is \[ \frac{1}{2} mv_0^2 - \frac{GMm}{R} \] At a distance \(r\), the total energy is \[ \frac{1}{2} mv^2 - \frac{GMm}{r} \]

Equalizing Initial and General Energy

To conserve energy, the initial energy should be equal to the energy at any distance \(r\): \[ \frac{1}{2} mv_0^2 - \frac{GMm}{R} = \frac{1}{2} mv^2 - \frac{GMm}{r} \]

Isolating the Velocity Term

Rearrange the equation to isolate the velocity \(v\): \[ \frac{1}{2} mv_0^2 + \frac{GMm}{r} - \frac{GMm}{R} = \frac{1}{2} mv^2 \] Dividing by \(m\) and simplifying: \[ v^2 = v_0^2 - \frac{2GM}{R} + \frac{2GM}{r} \]

Expressing Gravitational Potential

Recall that the acceleration due to gravity at Earth's surface is given by \( g = \frac{GM}{R^2} \). Thus, \( GM = gR^2 \). Substitute this in: \[ v^2 = v_0^2 - 2gR + \frac{2gR^2}{r} \]

Solving for Maximum Distance

For maximum distance \( r_{\text{max}} \), set \( v = 0\): \[ 0 = v_0^2 - 2gR + \frac{2gR^2}{r_{\text{max}}} \] Solving for \( r_{\text{max}} \): \[ r_{\text{max}} = \frac{2gR^2}{2gR - v_0^2} = \frac{R}{1 - \frac{v_0^2}{2gR}} \]

Finding Escape Velocity

For \( r \) to tend to infinity, the denominator must be zero: \[ 1 - \frac{v_0^2}{2gR} = 0 \] Therefore, the escape velocity \( v_e \) is: \[ v_e = \sqrt{2gR} \]

Key concepts

Gravitational force

Gravitational force is the attractive force that exists between any two masses. For an object of mass \(m\) at a distance \(r\) from the center of the Earth, the gravitational force follows an inverse square law and is given by:
\[ F = \frac{m g R^2}{r^2} \ onumber \] Here, \(g\) is the acceleration due to gravity at the Earth's surface and \(R\) is the Earth's radius. This force decreases with the square of the distance, meaning that as the object moves farther from the Earth, the force reduces significantly.

Differential equations

A differential equation involves functions and their derivatives, expressing how a function changes over time. In this exercise, the motion of a mass projected radially outward is described by:
\[ m \frac{d^2 r}{dt^2} = -m \frac{g R^2}{r^2} \] When the mass \(m\) is divided out, it simplifies to:
\[ \frac{d^2 r}{dt^2} = - \frac{g R^2}{r^2} \] This equation shows that the acceleration of the mass is inversely proportional to the square of the distance from the Earth's center.

Escape velocity

Escape velocity is the minimum velocity needed for an object to escape from a planet's gravitational field without any further propulsion. Derived from the conservation of energy, the escape velocity \(v_e\) from the Earth is given by:
\[ v_e = \frac{2gR^2}{2gR - v_0^2} \ onumber v_e = \text{simplified version} \sqrt{2gR} \ nonumber \] The escape velocity depends on the gravitational constant \(g\) and the Earth's radius \(R\). It's the speed at which the total energy (kinetic + potential) becomes zero, allowing the object to reach infinitely far distances.

Conservation of energy

The principle of conservation of energy states that the total energy of an isolated system remains constant. For the moving mass in our problem, the total energy includes kinetic and gravitational potential energy. Initially, at radius \(R\), the energy is:
\[ \frac{1}{2} mv_0^2 - \frac{GMm}{R} \] At a distance \(r\), the energy becomes:
\[ \frac{1}{2} mv^2 - \frac{GMm}{r} \] By equating these energies, we derive the relationship between the velocities and distances, allowing us to solve for different required quantities.

Kinetic and potential energy calculations

Kinetic energy and potential energy are two crucial components of mechanical energy in physics. Initially, the kinetic energy is:
\[ K.E. = \frac{1}{2} mv_0^2 \] The potential energy due to gravity at distance \(r\) is:
\[ U = - \frac{GMm}{r} \] Using the conservation of energy,
\[ \frac{1}{2} mv_0^2 - \frac{GMm}{R} = \frac{1}{2} mv^2 - \frac{GMm}{r} \] we can rearrange to solve for velocity \(v\) at any distance \(r\). This helps in finding the maximum distance and escape velocity, critical in gravitational motion studies.