True Weightlessness

Classical Mechanics Level 3

A spacecraft is travelling from Earth to the Moon in a straight-line path. As the spacecraft leaves the Earth towards the Moon, the magnitude of the gravitational pull by the Earth decreases and the magnitude of the gravitational pull by the Moon increases. At a point in between the Earth and the Moon, called the Lagrange point, the gravitational field strength becomes zero (due to cancelling out of gravitational pull) and hence the weight becomes zero. The mass of the Earth, the Moon and the distance between their centres, are given as:

$\begin{aligned} m_\oplus&=\SI{5.972 × 10^{24}}{\kg}\\ m_L&=\SI{7.345 × 10^{22}}{\kg}\\ \Delta_{\oplus L}&=\SI{3.844×10^8}{\m} \end{aligned}$

Find the distance (in $\ce{m}$ ) away from the centre of Earth at which the spacecraft experiences zero weight, i.e., the distance between the centre of the Earth and the Lagrange point.

Use:

$F=G\frac{m_1m_2}{r^2}$

The answer is 3.46E+8.

2 solutions

Gandoff Tan
Oct 9, 2019

$\text{Let the distance be }r.$

$\begin{aligned} F_\oplus&\overset{\text{set}}{=}F_L\\ G\frac{m_\oplus m}{r^2}&=G\frac{m_L m}{(\Delta_{\oplus L}-r)^2}\\ \frac{m_\oplus}{r^2}&=\frac{m_L}{(\Delta_{\oplus L}-r)^2}\\ \frac{(\Delta_{\oplus L}-r)^2}{r^2}&=\frac{m_L}{m_\oplus}\\ \left(\frac{\Delta_{\oplus L}-r}{r}\right)^2&=\frac{m_L}{m_\oplus}\\ \frac{\Delta_{\oplus L}}{r}-1&=\sqrt{\frac{m_L}{m_\oplus}}\\ \frac{\Delta_{\oplus L}}{r}&=1+\frac{\sqrt{m_L}}{\sqrt{m_\oplus}}\\ \frac{r}{\Delta_{\oplus L}}&=\frac{1}{1+\frac{\sqrt{m_L}}{\sqrt{m_\oplus}}}\\ r&=\left(\frac{\sqrt{m_\oplus}}{\sqrt{m_\oplus}+\sqrt{m_L}}\right)\Delta_{\oplus L} \end{aligned}$

$\begin{aligned} r&=\left(\frac{\sqrt{m_\oplus}}{\sqrt{m_\oplus}+\sqrt{m_L}}\right)\Delta_{\oplus L}\\ &=\left(\frac{\sqrt{5.972 × 10^{24}}}{\sqrt{5.972 × 10^{24}}+\sqrt{7.345 × 10^{22}}}\right)(3.844×10^8)\\ &\approx\boxed{\SI{3.46 \times 10^8}{\m}} \end{aligned}$

Debaparna Goswami
Nov 24, 2019

The centers of both bodies,i.e. the earth and the moon, are at a $\Delta_{\oplus L}$ distance apart, which is given as $\Delta_{\oplus L}= 3.844\times10^{8}$ m . At the Lagrange point, the gravitational pull on a mass from the two bodies must be at equilibrium, which means the gravitational potential of both the bodies at that point must be the same. Let $x$ be the distance of the Lagrange point from the earth and therefore, $\Delta_{\oplus L}-x$ is the distance of the Lagrange point from the moon.

Therefore, we can write,

$-\dfrac{Gm_{\oplus}}{x^{2}} = -\dfrac{Gm_{L}}{(\Delta_{\oplus L}-x)^{2}}$ ,

or, $\dfrac{5.972\times10^{24}}{x^{2}} = \dfrac{7.345\times10^{22}}{(\Delta_{\oplus L}-x)^{2}}$ (Putting the values of $m_{\oplus} and m_{L}$ )

or, $\dfrac{(\Delta_{\oplus L}-x)^{2}}{x^{2}} = \dfrac{7.345\times10^{22}}{5.972\times10^{24}}$

or, $\dfrac{(\Delta_{\oplus L}-x)^{2}}{x^{2}} = 1.23\times10^{-2}$

or, $\dfrac{(\Delta_{\oplus L}-x)}{x} = 0.110$

or, $\dfrac{\Delta_{\oplus L}}{x} = 1.110$

Solving for $x$ , we get,

$x = 3.463\times10^{8}$

True Weightlessness

The answer is 3.46E+8.

2 solutions

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