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Is centripetal force not always true for Planets?

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Source : IE Irodov: 1.207.

I used, force by gravitational pull of Sun should provide necessary centripetal force at one extreme $r_1$ where velocity is $v_1$ ( I took that moment as circle as $r_1\perp v_1$) ,$$\implies\dfrac{GM_sm}{r_1^2}=\dfrac{mv_1^2}{r_1}\implies v_1=\sqrt{\dfrac{GM_s}{r_1}}$$

But when I use energy conservation then, $$v_1=\sqrt{\dfrac{2GM_sr_2}{r_1(r_1+r_2)}}$$

Second expression gives me the correct answer, also if I put $r_1=r_2$ second expression reduces to first one, which says to me that we apply centripetal force method only when we have complete circle, not that instantaneous circles, why?

asked Sep 30 in Physics Problems by n3 (318 points)
edited Sep 30 by sammy gerbil

1 Answer

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Best answer

When $r_1=r_2=r$ the orbit is a circle of radius $r$. The gravitational force on the planet is then the centripetal force because the point to which the gravitational force on the planet is directed is the same as the centre of the circle in which it is moving.

However if $r_1 \ne r_2$ then the orbit is an ellipse not a circle. The local radius of curvature $r$ of the ellipse at each of the extremes $P_1, P_2$ (where $r_1, r_2$ are measured) is not $r_1, r_2$ respectively. See diagram below.

The gravitational forces at $P_1, P_2$ are directed towards the Sun $S$ and not towards the local centre of curvature $C$. To see this examine what happens when the planet has moved a short distance from $P_1$ to $P_1'$. Then the gravitational force is directed towards $S$ not $C$ - these are 2 different directions. This is true however close $P_1'$ is to $P_1$.

Therefore the gravitational force at $P_1, P_2$ is not equal to the centripetal force at these points.

answered Sep 30 by sammy gerbil (26,096 points)
selected Sep 30 by n3
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