91影视

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,$r=1.5脳{10}^{11}\mathrm{m}$

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

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

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

This problem is based on Newton鈥檚 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:

$F_s=\frac{G脳m_s脳m_p}{r_{sp}^2}$

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

$F_e=\frac{G脳m_e脳m_p}{r_{ep}^2}$

When both forces are at balance:

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

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,

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

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