A prototype of Jules Verne’s projectile, mass m, is launched vertically upwards from the Earth’s surface but only just reaches a height of one tenth of the Earth’s radius before falling back. When the height, s, above the surface is small compared with the radius of the Earth, R, the magnitude of

Question

A prototype of Jules Verne’s projectile, mass m, is launched vertically upwards from the Earth’s surface but only just reaches a height of one tenth of the Earth’s radius before falling back. When the height, s, above the surface is small compared with the radius of the Earth, R, the magnitude of the earth’s gravitational force on the projectile may be modelled as [latex]mg\left(1 - \frac{2s}{R}\right)[/latex], where g is gravitational acceleration at the Earth’s surface.

Assuming all other forces can be neglected
i) write down a differential equation of motion involving s and velocity, v
ii) integrate this equation and hence obtain an expression for the loss of kinetic energy of the projectile between its launch and rising to a height s
iii) show that the launch velocity is [latex]0.3\sqrt{2gR}[/latex].

Accepted Answer

i) Taking the upward direction as positive, the force on the projectile is [latex]-mg\left(1 - \frac{2s}{R}. ight)[/latex]

The force is a function of s, so start from the equation of motion in the form

[latex]mv\frac{dv}{ds} = - mg \left(1 - \frac{2s}{R}. ight).[/latex]

ii) Separating the variables and integrating gives

[latex] \begin{matrix} \int{mv dv} = - \int{mg\left(1 - \frac{2s}{R} ight)ds } & \boxed{ ext{You would normally divide the equation by m, but it is useful to leave it in here in order to get kinetic energy directly from} \int{mv dv} }\end{matrix} [/latex]

[latex]\Rightarrow \frac{1}{2}mv^{2} = -mgs + \frac{mgs^{2}}{R} + k[/latex]

Writing [latex]v_{0}[/latex] for the launch velocity v = [latex]v_{0}[/latex] when s = 0, so [latex]k = \frac{1}{2}mv_{0}^{2}[/latex] and rearranging gives

[latex]\frac{1}{2}mv_{0}^{2} - \frac{1}{2}mv^{2} = mgs - \frac{mgs^{2}}{R}.[/latex] ①

The left-hand side is the loss of kinetic energy, so

[latex] \begin{matrix} ext{loss of KE} = mgs - \frac{mgs^{2}}{R}. & \boxed{ ext{You can check that this is dimensionally consistent to give you confidence that your working is correct.} }\end{matrix} [/latex]

iii) Dividing equation ① by m and multiplying by 2 gives

[latex]v_{0}^{2} - v^{2} = 2gs - \frac{2gs^{2}}{R}.[/latex]

If the projectile just reaches a height [latex]s = \frac{R}{10},[/latex] then the velocity v is zero at that point.

Substituting [latex]s = \frac{R}{10}[/latex] and v = 0 gives

[latex]v_{0}^{2} = 2g\left(\frac{R}{10} ight) - \frac{2gR^{2}}{100R}[/latex]

[latex]= \frac{18gR}{100}[/latex]

⇒ [latex]v^{0} = \frac{3}{10} \sqrt{2gR}.[/latex]

So the launch velocity is [latex]0.3\sqrt{2gR}[/latex].

Question 18.2: A prototype of Jules Verne’s projectile, mass m, is launched...

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