Question

According to Newton's Law of Universal Gravitation, the gravitational force on an object of mass \(m\) that has been projected vertically upward from the earth's surface is

\(F = \frac{{mg{R^2}}}{{{{(x + R)}^2}}}\)

where \(x = x(t)\) is the object's distance above the surface at time t, R is the earth's radius, and \(g\) is the acceleration due to gravity. Also, by Newton's Second Law, \(F = ma = m(dv/dt)\) and so

\(m\frac{{dv}}{{dt}} = - \frac{{mg{R^2}}}{{{{(x + R)}^2}}}\)

(a) Suppose a rocket is fired vertically upward with an initial velocity \({v_0}\). Let \(h\) be the maximum height above the surface reached by the object. Show that

\({v_0} = \sqrt {\frac{{2gRh}}{{R + h}}} \)

(Hint: By the Chain Rule, \(m(dv/dt) = mv(dv/dx).)\)

(b) Calculate \({v_e} = \mathop {\lim }\limits_{h \to \infty } {v_0}\). This limit is called the escape velocity for the earth.

(c) Use \(R = 3960{\rm{mi}}\) and \(g = 32{\rm{ft}}/{{\rm{s}}^2}\) to calculate \({v_e}\) in feet per second and in miles per second.

Short answer

a)

The solution of given differential equation \(m\frac{{dv}}{{dt}} = - \frac{{mg{R^2}}}{{{{(x + R)}^2}}}\) is \({v_0} = \sqrt {\frac{{2gRh}}{{R + h}}} \).

b)

The escape velocity is \({v_e} = \mathop {\lim }\limits_{h \to \infty } {v_0} = \sqrt {2gR} \).

c)

The velocity is \(6.94{\rm{mi}}/{\rm{s}}\).

Step-by-step solution

Rule used

Chain Rule:

\(\begin{aligned}m\frac{{dv}}{{dt}} &= m\frac{{dv}}{{dx}} \cdot \frac{{dx}}{{dt}}\\m\frac{{dv}}{{dt}} &= mv\left( {\frac{{dv}}{{dx}}} \right)\end{aligned}\)

where, the object moved distance \(x\) in the surface at time \(t\) is called velocity.

Then, velocity \(v = \frac{{dx}}{{dt}}\).

Newton's Second Law: \(F = ma = m\left( {\frac{{dv}}{{dt}}} \right)\)

Solve the given differential equation

a)

The differential equation is

\(m\frac{{dv}}{{dt}} = - \frac{{mg{R^2}}}{{{{(x + R)}^2}}}\)

By using Chain rule,

\(\begin{aligned}m\frac{{dv}}{{dx}} \cdot \frac{{dx}}{{dt}} &= - frac{{mg{R^2}}}{{{{(x + R)}^2}}}\\v\frac{{dv}}{{dx}} &= - \frac{{g{R^2}}}{{{{(x + R)}^2}}}\end{aligned}\)

Separate the variables,

\(vdv = - \frac{{g{R^2}}}{{{{(x + R)}^2}}}dx\)

Integrate the above equation from both sides

Integrate on both sides,

\(\int v dv = - \int {\frac{{g{R^2}}}{{{{(x + R)}^2}}}} dx\)

\(\begin{aligned} &= - g{R^2}\int {\frac{1}{{{{(x + R)}^2}}}} dx\\ &= - g{R^2}\int {{{(x + R)}^{ - 2}}} dx\\\frac{{{v^2}}}{2} &= - g{R^2}\frac{{{{(x + R)}^{ - 2 + 1}}}}{{ - 2 + 1}} + C\end{aligned}\)

Simplify further,

\(\frac{{{v^2}}}{2} = g{R^2}{(x + R)^{ - 1}} + C\)

\(\frac{{{v^2}}}{2} = \frac{{g{R^2}}}{{(x + R)}} + C\)

Substitute the values in equation (1)

Substitute the initial condition \(v(0) = {v_0}\) in equation (1),

\(\begin{aligned}\frac{{v_0^2}}{2} &= \frac{{g{R^2}}}{{(0 + R)}} + C\\ &= \frac{{g{R^2}}}{R} + C\\ &= gR + C\\C &= \frac{{v_0^2}}{2} - gR\end{aligned}\)

Substitute \(C\) value in equation (1),

\(\frac{{{v^2}}}{2} = \frac{{g{R^2}}}{{(x + R)}} + \frac{{v_0^2}}{2} - gR\)

\(\frac{{{v^2}}}{2} - \frac{{v_0^2}}{2} = \frac{{g{R^2}}}{{(x + R)}} - gR\) ……. (2)

Given that \(h\) be the maximum height and that maximum height velocity will be zero.

That is \(v = 0\) and \(x = h\).

Substitute 0 for \(v\) and \(h\) for \(x\) in equation (2),

\(\begin{aligned}\frac{0}{2} - \frac{{{v_0}^2}}{2} &= \frac{{g{R^2}}}{{(h + R)}} - gR\\ - \frac{{{v_0}^2}}{2} &= \frac{{g{R^2}}}{{(h + R)}} - gR\\\frac{{{v_0}^2}}{2} &= gR - \frac{{g{R^2}}}{{(h + R)}}\\\frac{{v_0^2}}{2} = \frac{{ghR + g{R^2} - g{R^2}}}{{(h + R)}}\end{aligned}\)

Simplify further,

\(\begin{aligned}\frac{{v_0^2}}{2} &= \frac{{ghR}}{{(h + R)}}\\v_0^2 &= \frac{{2ghR}}{{(h + R)}}\end{aligned}\)

Take square root on both sides,

\({v_0} = \sqrt {\frac{{2ghR}}{{(h + R)}}} \)

Therefore, the solution of given differential equation \(m\frac{{dv}}{{dt}} = - \frac{{mg{R^2}}}{{{{(x + R)}^2}}}\) is \({v_0} = \sqrt {\frac{{2gRh}}{{R + h}}} \).

Hence proved.

Given information from question

b)

The maximum height goes to infinite.

\({v_e} = \mathop {\lim }\limits_{h \to \infty } {v_0}\)

Calculate the escape velocity

\(\begin{aligned}\mathop {\lim }\limits_{h \to \infty } {v_0} &= \mathop {\lim }\limits_{h \to \infty } \sqrt {\frac{{2gRh}}{{R + h}}} \\\mathop {\lim }\limits_{h \to \infty } {v_0} &= \mathop {\lim }\limits_{h \to \infty } \sqrt {\frac{{2gR}}{{\frac{R}{h} + 1}}} \\\mathop {\lim }\limits_{h \to \infty } {v_0} &= \sqrt {\frac{{2gR}}{{0 + 1}}} \\\mathop {\lim }\limits_{h \to \infty } {v_0} &= \sqrt {2gR} \end{aligned}\)

Therefore, the escape velocity is \({v_e} = \mathop {\lim }\limits_{h \to \infty } {v_0} = \sqrt {2gR} \).

Given information from question

c)

\(R = 3960\) miles and \(g = 32{\rm{ft}}/{{\rm{s}}^2}\).

Calculate the velocity

From part (b)

1 mile is equal to 5280 feet

So that 3960 miles / minute is 20908800 miles / minute.

\(\begin{aligned}{v_e} &= \mathop {\lim }\limits_{h \to \infty } {v_0} &= \sqrt {2(32){\rm{ft}}/{{\rm{s}}^2}(3960)(5280)} \\{v_e} &= \mathop {\lim }\limits_{h \to \infty } {v_0} &= \sqrt {64{\rm{ft}}/{{\rm{s}}^2}(20908800){\rm{feet}}/{\rm{ minute }}} \\{v_e} &= \mathop {\lim }\limits_{h \to \infty } {v_0} &= \sqrt {1338163200{\rm{feet}}/{\rm{second}}} \\{v_e} &= \mathop {\lim }\limits_{h \to \infty } {v_0} &= 36580{\rm{ft}}/s\end{aligned}\)

In miles, 1 feet per second is 0.000189 miles per second

Therefore, \(36580{\rm{ft}}/{\rm{s}}\) is \(6.94{\rm{mi}}/{\rm{s}}\).

Thus, the velocity is \(6.94{\rm{mi}}/{\rm{s}}\).