Question

Infinite number of masses, each of mass 3 kg are
placed along the y-axis at y = 1 m, 3 m, 9 m,
27 m ... The magnitude of resultant gravitational
potential in terms of gravitational constant G at the
origin (y = 0) is
(1) 4.5 G unit (2) G unit
(3) 3 G unit
(4) 9 G unit​

Answer 1

ANSWER:

The resultant gravitational potential energy is 4.5 G unit.

EXPLANATION:

The formula for gravitational potential energy is  

$V=\frac{G M}{R}$

As the gravitational potential energy is dependent on mass M and distance R where the mass M is kept from the reference point, the resultant gravitational potential energy will the sum of the gravitational potential energy exhibited by each mass at different positions.

$V=\sum_{i=1}^{\infty} V_{i}$

So the resultant gravitational potential energy acting at the reference point is  

$V=\sum_{i=1}^{\infty} \frac{G M_{i}}{R_{i}}$

Since the mass is same for all the masses attached, so we can take the values G and M as common,

$V=G M \sum_{i=1}^{\infty} \frac{1}{R_{i}}$

$V=3 G\left[\frac{1}{1}+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\cdots\right]$

As the values inside the square bracket are in a geometric progression with their starting value as 1 and r = 1/3 and n = 0 to infinitity. The sum of infinite terms of geometric progression (G.P) series is as follows

$S_{\infty}=\frac{a}{1-r}$

Here “a” is the starting value of the G.P series and r is the common ratio. So in this case, a =1 and r = 1/3.

$V=3 G\left[\frac{1}{1-\left(\frac{1}{3}\right)}\right]=3 G\left[\frac{1}{2 / 3}\right]$

$V=3 G * \frac{3}{2}=\frac{9 G}{2}=4.5 G \text { units }$

Thus, the resultant gravitational potential energy is 4.5 G unit.

Answer 2

Answer:

hope this helps you !!!

pls, mark as brainliest  by pressing the crown icon below   ^_^