The minimum and maximum distances of a satellite from the centre of earth are2Rand 4Rrespectively, whereRis the radius of earth and Mis the mass of the earth. The radius of curvature at the point of minimum distance is λR. Find the value ofλ.Correct answer is '2.667'. Can you explain this answer?

Question

The minimum and maximum distances of a satellite from the centre of earth are2Rand 4Rrespectively, whereRis the radius of earth and Mis the mass of the earth. The radius of curvature at the point of minimum distance is &lambda;R. Find the value of&lambda;.Correct answer is '2.667'. Can you explain this answer?

Answer

To find the radius of curvature at the point of minimum distance, we can use the formula for the radius of curvature of a curve:

$$R = \frac{{(1 + y'^2)^{3/2}}}{{y''}}$$

where $y$ is the function describing the curve, $y'$ is the first derivative of $y$ with respect to $x$, and $y''$ is the second derivative of $y$ with respect to $x$.

In this case, we can consider the satellite's distance from the center of the Earth as the function $y(x)$, where $x$ is the angle around the Earth's center. Since the distance varies between 2R and 4R, we can write the function as:

$y(x) = 2R + (4R - 2R)\sin(x)$

Taking the derivative of $y(x)$ with respect to $x$, we get:

$y'(x) = (4R - 2R)\cos(x)$

And taking the derivative of $y'(x)$ with respect to $x$, we get:

$y''(x) = -(4R - 2R)\sin(x)$

Plugging these values into the formula for the radius of curvature, we have:

$$R = \frac{{(1 + (4R - 2R)^2\cos^2(x))^{3/2}}}{{-(4R - 2R)\sin(x)}}$$

Since we want to find the radius of curvature at the point of minimum distance, we need to find the value of $x$ that corresponds to this point. The point of minimum distance is at the bottom of the satellite's trajectory, which corresponds to the angle $x = \pi$.

Plugging in $x = \pi$ into the formula, we get:

$$R = \frac{{(1 + (4R - 2R)^2\cos^2(\pi))^{3/2}}}{{-(4R - 2R)\sin(\pi)}}$$

Simplifying, we have:

$$R = \frac{{(1 + (2R)^2(-1))^{\frac{3}{2}}}}{{-(2R)\cdot 0}}$$

Since $\sin(\pi) = 0$, the denominator becomes zero, which means the radius of curvature at the point of minimum distance is undefined.

Answer

Applying conservation of angular momentum

From conservation of energy

Solving Eqs. (1) and (2), we get

If r is the radius of curvature at point A

The correct answer is: 2.667

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