Two astronauts are floating in gravitation-free space after having lost contact with their spaceship. The two will:

Two particles of mass \(m\) and \(4m\) are separated by a distance \(r.\) Their neutral point is at:
1. \(\frac{r}{2}~\text{from}~m\)
2. \(\frac{r}{3}~\text{from}~4m\)
3. \(\frac{r}{3}~\text{from}~m\)
4. \(\frac{r}{4}~\text{from}~4m\)

Three identical point masses, each of mass \(1~\text{kg}\) lie at three points \((0,0),\)  \((0,0.2~\text{m}),\)  \((0.2~\text{m}, 0).\) The net gravitational force on the mass at the origin is:
1. \(6.67\times 10^{-9}(\hat i +\hat j)~\text{N}\)
2. \(1.67\times 10^{-9}(\hat i +\hat j) ~\text{N}\)
3. \(1.67\times 10^{-9}(\hat i -\hat j) ~\text{N}\)
4. \(1.67\times 10^{-9}(-\hat i -\hat j) ~\text{N}\)

Two spheres of masses \(m\) and \(M\) are situated in air and the gravitational force between them is \(F.\) If the space around the masses is filled with a liquid of specific density \(3,\) the gravitational force will become:
1. \(3F\)
2. \(F\)
3. \(F/3\)
4. \(F/9\)

A planet of mass \(m\) is moving around a star of mass \(M\) and radius \(R\) in a circular orbit of radius \(r.\) The star abruptly shrinks to half its radius without any loss of mass. What change will be there in the orbit of the planet?

The law of gravitation states that the gravitational force between two bodies of mass \(m_1\) $\mathrm{and}$ \(m_2\) is given by:
\(F=\dfrac{Gm_1m_2}{r^2}\)
$\mathrm{}$\(G\) (gravitational constant) \(=7\times 10^{-11}~\text{N-m}^2\text{kg}^{-2}\)$\mathrm{.}$
\(r\) (distance between the two bodies) in the case of the Earth and Moon $\mathrm{}$\(=4\times 10^8~\text{m}\)
\(m_1~(\text{Earth})=6\times 10^{24}~\text{kg}\)
\(m_2~(\text{Moon})=7\times 10^{22}~\text{kg}\)
${\mathrm{}}_{}$What is the gravitational force between the Earth and the Moon?
1. \(1.8375 \times 10^{19}~\text{N}\)
2. \(1.8375 \times 10^{20}~\text{N}\)
3. \(1.8375 \times 10^{25}~\text{N}\)
4. \(1.8375 \times 10^{26}~\text{N}\)