Question

Two spherical bodies of mass $M$ and $5\;M$ and radii $R$ and $2\;R$ are released in free space with initial separation between their centres equal to $12\;R$. If they attract each other due to gravitational forces only, then the distance covered by the smaller body before collision is:

& A.7.5R \\\ & B.1.5R \\\ & C.2.5R \\\ & D.4.5R \\\ \end{aligned}$$

Answer 1

The gravitational force between the centres of the two balls is given by the law of gravitation. From that equation we can find the relationship between the force and radius of the balls. Using the given information, we can find the unknown.

Formula used:
$F=G\dfrac{m_{1}m_{2}}{r^{2}}$

Complete step-by-step answer:
We know that the gravitational force is a non-contact force. We know that according to Newton- Kepler law of gravitation, gravitational force is given as, the force due to gravitation is given as $F=G\dfrac{m_{1}m_{2}}{r^{2}}$, where $G$ is the gravitational constant, $m_{1},m_{2}$ is the masses of the body, and $r$ is the distance between the two bodies.
Given that the spherical bodies of mass $5\;M$ and $M$ with radii $2\;R$ and $R$, the distance between the bodies is $12R-R-2R=9R$
Let us assume, that the bodies move towards each other by a distance $s_{1}$ and $s_{2}$ then, $s_{1}+s_{2}=9R$
From newton’s second law, we know that $F=ma$ or $a\propto \dfrac{1}{m}$
Then, we have $\dfrac{a_{1}}{a_{2}}=\dfrac{5M}{M}$
$\implies \dfrac{a_{1}}{a_{2}}=5$
If both the bodies are at rest initially, we have$u_{1}=u_{2}=0$
Then, from $s=ut+\dfrac{1}{2}at^{2}$
We can say that, $s_{1}=\dfrac{1}{2}a_{1}t^{2}$
And similarly, $s_{2}=\dfrac{1}{2}a_{2}t^{2}$
Since the duration is same for both the bodies, we have $\dfrac{s_{1}}{s_{2}}=\dfrac{a_{1}}{a_{2}}$
$\implies \dfrac{s_{1}}{s_{2}}=5$
$\implies s_{1}=5s_{2}$
$\implies 5s_{2}+s_{2}=9R$
$\implies 6s_{2}=9R$
$\implies s_{2}=1.5R$
Here, we have taken $s_{2}$ to be the distance travelled due to the small body. Thus the correct answer is option $B.1.5\;R$

So, the correct answer is “Option B”.

Additional Information: The acceleration due to gravitation on the surface of the earth is given as $g=\dfrac{Gm}{R^{2}}$ also, this value changes with respect to the height of the object from the centre of the earth. This force is generally attractive in nature.

Note: Generally, $M$ and $m$ is the mass of the bigger and the smaller objects which are at a distance $r$ with respect to each other. Also note that, $m$ is the mass of the small object which is attracted to a bigger object of mass $M$. Since the $M >> m$, we are considering the effect of the small body for finding the field.