-1
$\begingroup$

Here's a question that's creating some doubt to me.

Suppose there are 2 big spheres A and B of mass M and mass 4M, each of radius, R separated by a distance of 6R.
An object of mass, m is projected from the surface of A. What should be the minimum velocity of the body with which it should be projected so that it just reaches the surface of B.

Figure of the question

A try to the question:-

We'll first find the neutral point where the gravitational forces by both the objects cancel out each other.
For this, we give a little displacement to the object/satellite, $d\vec{r}$ which is in direction from A to B.
Also, a unit vector $\hat{r}$ is assigned in the same direction.
Suppose the gravitational forces of the masses be $\vec{F_A}$ and $\vec{F_B}$. They are given below:

$\vec{F_A}$ = - $G\frac{Mm}{r^2}\hat{r}$
$\vec{F_B}$ = $G\frac{4Mm}{x^2}\hat{r}$

Negative sign is not present in $\vec{F_B}$ because $\hat{r}$ is in the direction of the force $\vec{F_B}$.

Now, at neutral point, P

$\vec{F_A}$ = - $\vec{F_B}$
FA = FB

$G\frac{Mm}{r^2}$ = $G\frac{4Mm}{x^2}$ [Here, x=6R-r]
$G\frac{Mm}{r^2}$ = $G\frac{4Mm}{(6R-r)^2}$

4 r2 = (6R-r)2

r = 2R

From this point, P(r=2R), the gravitational force, FB is sufficient to attract the satellite to reach the surface of B.

Let WA and WB be the work done by the gravitational forces $\vec{F_A}$ and $\vec{F_B}$ separately from the surface of A to point P.


Work done by Force, FA:-

dWA = $\vec{F_A}\cdot d\vec{r}$
dWA = FA dr cos 180°

dWA = - FA dr ---------Eq(a)

In equation a,

For limits,
Now, when object is at surface of A, r = R
And, when object is at neutral point P, r = 2R

$$\int \, dW_A = \int\limits_{R}^{2R} - F_A \, dr$$
$$W_A = - \int\limits_{R}^{2R} G\frac{Mm}{r^2} \, dr$$
$$W_A = -GMm \int\limits_{R}^{2R} \frac{1}{r^2} \, dr$$ $$W_A = -GMm \biggl[\frac{-1}{r}\biggr]_{R}^{2R} $$
$$W_A = -GMm \biggl[\frac{-1}{2R}-\frac{-1}{R}\biggr] $$
$$W_A = -\frac{GMm}{2R} $$

$$W_A = {\color{violet}{\int\limits_{R}^{2R} - F_A \, dr}} = {\color{pink}{-\frac{GMm}{2R}}} $$

Both equations, violet and the pink one are satisfying each other, infering that the work done by the gravitational force $\vec{F_A}$ will be negative.


Work done by Force, FB:-

dWB = $\vec{F_B}\cdot d\vec{r}$
dWB = FB dr cos 0°

dWB = FB dr ---------Eq(b)

In equation b,

For limits,
Now, when object is at surface of A, r = R -----> x = 6R - r = 5R
And, when object is at neutral point P, r = 2R -----> x = 6R - r = 4R

$$\int \, dW_B = \int\limits_{R}^{2R} F_B \, dr$$
$$W_B = \int\limits_{R}^{2R} G\frac{4Mm}{(6R - r)^2} \, dr$$
$$W_B = 4GMm\int\limits_{5R}^{4R} \frac{1}{x^2} \, dr$$
$$W_B = 4GMm \biggl[\frac{-1}{x} \biggr]_{5R}^{4R} $$ $$W_B = 4GMm \biggl[\frac{-1}{4R}-\frac{-1}{5R}\biggr] $$ $$W_B = 4GMm \biggl[\frac{-1}{20R} \biggr] $$ $$W_B = -\frac{GMm}{5R} $$

$$W_B = {\color{orange}{\int\limits_{R}^{2R} F_B \, dr}} = {\color{cyan}{-\frac{GMm}{5R}}}$$

So, here's my doubt:-

The orange equation infer that the work by gravitational force $\vec{F_B}$ is positive (why) as it was derived from equation (b) and in that equation, the angle between force $\vec{F_B}$ and displacement $d\vec{r}$ was 0°. And, it is equal to the cyan equation.
But in the cyan equation, there is a negative sign which tells us that the work by gravitational force $\vec{F_B}$ is negative.

Hence, the orange equation is not consistent with cyan equation.

My Doubt: Why?

Well, there's lot to be done as we need to find the velocity of satellite with which it should be projected. But, before that, I need to add the 2 work, WA and WB and make it equal to the kinetic energy of the satellite to find the velocity of the body (if I'm not wrong).

But, I'm stuck here. So please help.

OK, I have done a lot in the post but the doubt is lot like similar to the last one asked in this post Work Done by Gravitational Force.
The difference is only that in that post, I had some problem associated to the direction of radial vector $d\vec{r}$. But here, I don't think so, there's any problem with that.

So, please tell me why the the orange equation is not consistent with cyan equation?

$\endgroup$

closed as off-topic by John Rennie, Kyle Kanos, Jon Custer, stafusa, JMac Nov 14 '17 at 19:19

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – John Rennie, Kyle Kanos, Jon Custer, stafusa, JMac
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ If you are using $x=6R-r$ in form two, shouldn't it be $dx=-dr$? $\endgroup$ – Kyle Kanos Nov 13 '17 at 11:01
  • $\begingroup$ @Kyle Kanos, please refer to the comment given in eranreches answer. $\endgroup$ – lakhi Nov 13 '17 at 18:05
1
$\begingroup$

In your transition between the following two equations

$$W_B = \int\limits_{R}^{2R} G\frac{4Mm}{(6R - r)^2} \, dr$$
$$W_B = 4GMm\int\limits_{5R}^{4R} \frac{1}{x^2} \, dr$$

you've forgotten a minus sign due to $dr=-dx$.

$\endgroup$
  • $\begingroup$ yeah, you are right. I made a mistake. But if you look, $d\vec{r}$ = $-d\vec{x}$. Both are vectors, equal and opposite in direction . So, dr cannot be equal to -dx because both are length vector and length cannot have a negative value. $\endgroup$ – lakhi Nov 13 '17 at 18:04
  • 1
    $\begingroup$ $dx$ isn't simply length. It is the differential. Loosely speaking, when $r$ increases ($dr>0$) $x$ decreases ($dx<0$), since $x=6R-r$, and thus $dr=-dx$. $\endgroup$ – eranreches Nov 13 '17 at 18:07
  • $\begingroup$ so sorry. I completely ignored this fact. Just now, it hits my head. Thanks for sorting me out in both the posts. Vlauable answer. Thanks!!!! $\endgroup$ – lakhi Nov 13 '17 at 18:26

Not the answer you're looking for? Browse other questions tagged or ask your own question.