We want to position a space probe along a linethat extends directly toward the Sun in order to monitor solar flares.How far from Earth’s center is the point on the line where the Sun’sgravitational pull on the probe balances Earth’s pull?

i) Distance from the center of the Sun to probe$=r_{sp}$

ii) Distance from the center of the Earth to probe$=r_{ep}$

iii) Distance between the center of the Earth to the center of the Sun,$r=1.5\times {10}^{11}\mathrm{m}$

iv) Mass of the Earth,$m_e=5.98\times {10}^{24}\mathrm{kg}$

v) Mass of the Sun,$m_s=1.99\times {10}^{30}\mathrm{kg}$

$where,r_{sp}=r-r_{ep}$

This problem is based on Newton’s law of gravitation. The distance between sun and probe can be found by using the distance between center of Sun and Earth and the distance between center of Earth and probe.

Formula:

The gravitational force of attraction, $F=\frac{GMm}{r^2}$ (i)

Using equation (i), Force due to Sun on the probe can be given as:

$F_s=\frac{G\times m_s\times m_p}{r_{sp}^2}$

Using equation (i), Force due to Earth on the probe can be given as:

$F_e=\frac{G\times m_e\times m_p}{r_{ep}^2}$

When both forces are at balance:

$$\begin{gathered} F_s=F_e \\ \frac{G\times m_s\times m_p}{r_{sp}^2}=\frac{G\times m_e\times m_p}{r_{ep}^2} \end{gathered}$$

Hence,

$\frac{m_s}{r_{sp}^2}=\frac{m_e}{r_{ep}^2}$

But,$r_{sp}=r-r_{ep}$

Hence, equation (ii) can be written as,

$$\begin{gathered} \frac{m_s}{{\left(r-r_{ep}\right)}^2}=\frac{m_e}{r_{ep}^2} \\ \frac{1.99\times {10}^{30}\mathrm{kg}}{{\left(1.5\times {10}^{11}\mathrm{m}-{\mathrm{r}}_{\mathrm{ep}}\right)}^2}=\frac{5.98\times {10}^{24}\mathrm{kg}}{r_{ep}^2} \\ {r}_{ep}^2=3.005\times {10}^{-6}{\left(1.5\times {10}^{11}\mathrm{m}-{\mathrm{r}}_{\mathrm{ep}}\right)}^2 \\ {r}_{ep}=1.734\times {10}^{-3}{\left(1.5\times {10}^{11}\mathrm{m}-{\mathrm{r}}_{\mathrm{ep}}\right)}^2 \\ {r}_{ep}=2.601\times {10}^8-1.734\times {10}^{-11}{\mathrm{r}}_{\mathrm{ep}} \\ \approx 2.601\times {10}^8\mathrm{m} \end{gathered}$$

Hence, the probe is at a distance of $2.601\times {10}^8\mathrm{m}$from the center of the Earth.