Q.Supposing Newton's law of gravitation for gravitation forces $\mathbf{F}_1$ and $\mathbf{F}_2$ between two masses $m_1$ and $m_2$ at positions $\mathbf{r}_1$ and $\mathbf{r}_2$ read $\mathbf{F}_1 = -\mathbf{F}_2 = -\dfrac{\mathbf{r}_{12}}{r_{12}^{3}}\, G M_0^{2} \left(\dfrac{m_1 m_2}{M_0^{2}}\right)^{n}$ where $M_0$ is a constant of dimension of mass, $\mathbf{r}_{12} = \mathbf{r}_1 - \mathbf{r}_2$ and $n$ is a number. In such a case, (Note: more than one of the given options may be correct.)

Setting up

The modified force law is

In the usual case $n=1$, this reduces to Newton's law, and the force is directly proportional to $m_1m_2$. For $n\neq1$, the force instead scales as $(m_1m_2)^n$ — a genuinely different dependence on mass.

(A) Acceleration due to gravity on Earth

For an object of mass $m$ on Earth's surface (mass $M_E$, radius $R_E$), the force magnitude is

so the acceleration is

Unless $n=1$, this depends on the object's own mass $m$ — different objects would fall with different accelerations. (A) is correct.

(B) and (C) Kepler's laws

Kepler's first law (elliptical orbits) and second law (equal areas in equal times) both follow from the force being central and varying purely with distance as $1/r^2$ — neither property has changed here, only the mass-dependence has. So the first and second laws still hold.

Kepler's third law, however, comes from equating the force to the centripetal requirement for a planet of mass $m$ orbiting a star of mass $M$:

$m\frac{4\pi^2}{T^2}r = \frac{GM_0^2}{r^2}\left(\frac{Mm}{M_0^2}\right)^n$ …

Step 1: F=GM0^(2-2n)(Me*m)^n/r^2, so a=F/m is proportional to m^(n-1). Step 2: a depends on m unless n=1 -- different objects accelerate differently -- (a) true. Step 3: Kepler I/II depend only on the force being central and 1/r^2 -- still hold; Kepler III relies on cancellation of orbiting mass, which needs n=1 -- fails for n!=1, only the third law breaks -- (c) true, (b) false. Ste …