i) Denoting the Earth’s radius by R metres, show that the gravitational force on a body of mass m kg above the Earth’s surface and a distance s metres from the Earth’s centre (s R) is mgR²/s² . ii) A projectile is fired vertically from the Earth’s surface with an initial velocity of u ms^-1 and

Question

i) Denoting the Earth’s radius by R metres, show that the gravitational force on a body of mass m kg above the Earth’s surface and a distance s metres from the Earth’s centre (s > R) is [latex]\frac{mgR^{2}}{s^{2}}.[/latex]
ii) A projectile is fired vertically from the Earth’s surface with an initial velocity of u [latex]ms^{-1}[/latex] and reaches a maximum height of h m.
Derive from Newton’s second law an expression giving u² in terms of R, g and h.
(Neglect air resistance.)
iii) For what launch speed would the projectile just reach a height equal to the radius of the Earth, 6400 km? (Use g = 9.8 [latex]ms^{-2}[/latex].)
iv) What is the minimum launch speed if the projectile is never to return?

Accepted Answer

i) At the Earth’s surface, the force on a body of mass m is [latex]\frac{GMm}{R^{2}}[/latex] newtons. So

[latex]\frac{GMm}{R^{2}} = mass × acceleration = mg[/latex]

⇒ GM = gR².

Above the Earth’s surface at a distance s metres from the centre, the force on a body of mass m is

[latex]\frac{GMm}{s^{2}} = \frac{gR^{2}m}{s^{2}}[/latex] (substituting GM = gR²)

= [latex]\frac{mgR^{2}}{s^{2}}[/latex].

ii) Take the positive direction (increasing s) upwards. The force is acting downwards, hence the negative sign when applying the equation of motion.

[latex]- \frac{mgR^{2}}{s^{2}} = mv\frac{dv}{ds}[/latex]

The [latex]v\frac{dv}{ds}[/latex] form for acceleration is used since the problem involves distances and velocities.

Dividing by m and separating the variables gives

[latex]\int{v dv} = - \int{\frac{gR^{2}}{s^{2}}} ds[/latex]

⇒ [latex]\frac{1}{2}v^{2} = \frac{gR^{2}}{s} + c.[/latex]

s is the distance from the centre of the Earth, so v = u when s = R, giving [latex]c = \frac{1}{2}u^{2} - gR[/latex],

⇒ [latex]\frac{1}{2}v^{2} = \frac{gR^{2}}{s} + \frac{1}{2}u^{2} - gR[/latex],

v = 0 when s = h + R gives

[latex]\frac{1}{2}u^{2} = gR - \frac{gR^{2}}{h + R}[/latex] ①

[latex]u^{2} = \frac{2gRh}{h + R}.[/latex]

iii) When h = R, u² = gR

⇒ [latex] \begin{matrix} u = \sqrt{gR} = 7920. & &\longleftarrow \boxed{R = 6400 imes 10^{3} ext{ in metres.}} \end{matrix} [/latex]

The launch speed is approximately 7.9 km[latex]s^{-1}.[/latex]

iv) [latex]\frac{gR^{2}}{h + g}[/latex] → 0 as h → ∞ so equation ① gives [latex]\frac{1}{2}u2 → gR[/latex] for large h.

In order never to return, the minimum launch speed is [latex]\sqrt{2gR} = 11 200 ms^{-1}.[/latex]

This is 11.2km[latex]s^{-1}.[/latex] For obvious reasons, this value of u is known as the escape velocity.
An alternative solution using energy methods is given in Example 18.6 on pages 415 to 416.

Question 18.4: i) Denoting the Earth’s radius by R metres, show that the gr...

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