Two particles with masses m are placed at the (0, a) and (0,...

Two particles with masses m are placed at the (0, a) and (0, -a) points on y-axis. Find the magnitude of gravitational acceleration (g) at the point P(x,0) on x-axis. \begin{enumerate} \item $\frac{4Gmx}{(x^2+a^2)^{1/2}}$ \item 0 \item $\frac{2Gmx}{(x^2+a^2)^{3/2}}$ \item $\frac{4Gmx}{(x^2+a^2)^{3/2}}$ \item $\frac{2Gmx}{(x^2+a^2)^{1/2}}$ \end{enumerate}

Transcript

00:01 So this is essentially a diagram of what's going on.

00:03 We can say that the net gravitational field will be equal to gravity to the right plus the gravity to the left.

00:18 You may notice that if this angle is theta and this angle is theta in the in the x direction, they're actually going to cancel out.

00:32 So we actually don't have to find the x direction.

00:40 In fact, here, because based on symmetry, we know that the x direction is going to cancel out.

00:46 We can actually just simply find the magnitude of 1 in the, magnitude of 1 in the y direction, and then multiply that by 2, given that the, we can say that g is simitimate.

01:03 Equal to g right rather we can say that the gravity on the right is equal to the y component of the gravity of the y will equal the the y component on the gravity on the left and we can say that the x component will be opposite of one another so this is just based on symmetry so that we don't have to write so many terms and at that point we can just say that uh uh g is simply going to be equal to, in this case it'll be negative 2gm over x not squared plus y not squared, cosine of theta times j hat.

02:11 Now, i said negative because we're going down.

02:16 And if we wanted to substitute for cosine of theta, we can say that g, the vector, the gravity vector is going to be equal to negative 2gm over x not squared plus y not squared times y divided by the square root of x not squared plus y not squared and so again j hat and so this will equal negative 2 g m y divided by x divided by x and not squared plus y not squared to the three halves power times j hat.

03:07 So this would be our final answer.

03:14 And so for part b, when they're asking for the maximum acceleration due to gravity, we can say that g equals 2g and y divided by.

03:30 So essentially we're taking a relationship from part a, and we're saying dg with respect to time, will equal 2gm.

03:43 And then here is where we have to make the substitution.

03:49 And let's say dg over dt equals 2gm.

03:55 And then here it'll be x not squared plus y not squared to the 3 half power minus y times 3 over 2 times x not squared plus y not squared to the one half power times.

04:18 Times 2y and this will be divided by x not squared plus y squared to the third power.

04:34 Oh my apologies actually these are the subscript actually goes away and so at this point we can say that this is going to equal zero...

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