Need help with Simple Harmonic Question (1 Viewer)

max_ma · Oct 4, 2013

An object falling directly on the earth from space moves according to the equation (d^2 x)/(dt^2 )= (-k)/x^2 , where x is the distance of the object from the centre of the earth at time t.

The constant k is related to g, the value of gravity at the earth's surface, and the radius of the earth, R by the formula k = gR^2.

Show that, if the object started from rest at a distance of 10^9 metres from the earth's centre, then it will reach earth with a velocity of approximately 11 200 m/s.

[Assume R = 6.4 x 10^6m, g - 9,8 m/s^2]

Thanks

ymcaec · Oct 4, 2013

max_ma said:
An object falling directly on the earth from space moves according to the equation (d^2 x)/(dt^2 )= (-k)/x^2 , where x is the distance of the object from the centre of the earth at time t.

The constant k is related to g, the value of gravity at the earth's surface, and the radius of the earth, R by the formula k = gR^2.

Show that, if the object started from rest at a distance of 10^9 metres from the earth's centre, then it will reach earth with a velocity of approximately 11 200 m/s.

[Assume R = 6.4 x 10^6m, g - 9,8 m/s^2]

Thanks

$\frac{\mathrm{d} ^2x}{\mathrm{d} t^2}=-\frac{k}{x^2}\\\frac{\mathrm{d} }{\mathrm{d} x}\left ( \frac{1}{2} v^2 \right )=-\frac{k}{x^2}\\\frac{1}{2} v^2 =\frac{k}{x} + c\\\\$When $x$ = $10^9 $ m, $v=0$:$\\c = -\frac{k}{10^9}\\\\$When $x$ = $6.4\times 10^6 $ m, $v$:$\\\frac{1}{2} v^2 =\frac{k}{6.4\times 10^6} -\frac{k}{10^9}\\$And since $k = gR^2 = 9.8 \times (6.4 \times 10^6)^2 = 4.01... \times 10^{14}\\v=11164.10.... \approx 11 200$ m/s$$

and btw not SHM

Need help with Simple Harmonic Question (1 Viewer)

max_ma

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ymcaec

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