Answer
At a distance of $~~2.59\times 10^8~m~~$ from Earth’s center is the point on the
line where the Sun’s gravitational pull on the probe balances Earth’s pull.
Work Step by Step
The gravitational force on the probe due to the sun is directed toward the sun.
The gravitational force on the probe due to the Earth is directed toward the Earth.
For the net gravitational force on the probe to be zero, the gravitational force on the probe due to the sun must be equal in magnitude to the gravitational force on the probe due to the Earth.
Let $r$ be the distance from the center of the Earth to the probe.
Then $~~(1.5\times 10^{11}~m-r)~~$ is the distance from the sun to the probe.
We can find $r$:
$\frac{G~m_E~m_P}{r^2} = \frac{G~m_S~m_P}{(1.5\times 10^{11}-r)^2}$
$m_E~(1.5\times 10^{11}-r)^2 = m_S~r^2$
$\sqrt{m_E}~(1.5\times 10^{11}-r) = \sqrt{m_S}~r$
$(\sqrt{m_S}+\sqrt{m_E})~r = \sqrt{m_E}~(1.5\times 10^{11})$
$r = \frac{\sqrt{m_E}~(1.5\times 10^{11})}{\sqrt{m_S}+\sqrt{m_E}}$
$r = \frac{\sqrt{5.98\times 10^{24}}~(1.5\times 10^{11})}{\sqrt{2.0\times 10^{30}}+\sqrt{5.98\times 10^{24}}}$
$r = (0.001726)(1.5\times 10^{11}~m)$
$r = 2.59\times 10^8~m$
At a distance of $~~2.59\times 10^8~m~~$ from Earth’s center is the point on the
line where the Sun’s gravitational pull on the probe balances Earth’s pull.
