08_একাদশ শ্রেণী_পদার্থবিদ্যা_মহাকর্ষ - পৃথিবীর অক্ষীয় ঘূর্ণনের জন্য অভিকর্ষজ ত্বরণের মানের পরিবর্তন
08_একাদশ শ্রেণী_পদার্থবিদ্যা_মহাকর্ষ - পৃথিবীর অক্ষীয় ঘূর্ণনের জন্য অভিকর্ষজ ত্বরণের মানের পরিবর্তন
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LearningTranscript
00:00We know that the Earth is constantly revolving around its axis, which is why it is day and night on Earth.
00:16You know that an uncertain ball is formed due to the motion of the Earth.
00:21An object that is not present in the universe of a revolving ball can slip out of the ball due to this uncertain ball.
00:28You may be wondering that the Earth is also a ball.
00:31Even if it is not a completely geometric ball, it is almost a ball.
00:36So, why are the objects in the universe, i.e. you, me, bus, train, etc., not slipping out of the universe due to the intense rotation of the Earth?
00:48It is because of the uncertain ball.
00:52Fortunately, the massive ball of the Earth is much stronger than the uncertain ball.
00:58But is there no effect of the uncertain ball created for this rotation?
01:03Yes, there is.
01:05This uncertain ball generates a lot of heat in the atmosphere.
01:10It is not difficult to understand that the weight of the object in the atmosphere is actually the weight of this heat.
01:17Will this amount of heat be equal everywhere in the atmosphere?
01:20Let's find out.
01:23In the past, our Earth has revolved around itself in the form of an omega cone.
01:29The circumference of the Earth is capital R.
01:31Therefore, the uncertain ball of a small m-sized object in the D point of the universe will be N into omega square R.
01:41Therefore, the weight of the object will be mg' is equal to mg minus M into omega square R.
01:50Now, if we take the same object in the direction of theta,
01:55the uncertain ball of the object will be M into omega square into small r.
02:02But the entire weight of this uncertain ball will not be affected.
02:07Because the massive ball of the object will work in the same direction.
02:13Therefore, the massive ball of the object will work in the direction of M into omega square small r cos theta.
02:21How did I say this?
02:23Let's see the picture again.
02:25The object is now in the direction of theta.
02:28That is, the theta of the cone will be in the direction of A and in the direction of O.
02:32Therefore, the massive ball of the object will work in the direction of M into omega square small r cos theta.
02:39Therefore, the massive ball of the object will work in the direction of A and in the direction of O.
02:44Therefore, the weight of the object will be mg' is equal to mg minus M into omega square small r cos theta.
02:53Again, its value is capital R.
02:56Therefore, the value of small r will be capital R cos theta.
03:00Similarly, the value of small r will be capital R cos theta.
03:04Therefore, mg' is equal to mg minus M into omega square capital R cos square theta.
03:11Therefore, g' is equal to g minus omega square capital R cos square theta.
03:18Therefore, g' is equal to g into within bracket 1 minus omega square capital R cos square theta by g.
03:28In other words, g minus g' is equal to del g is equal to omega square capital R cos square theta.
03:39That is, del g is proportional to cos square theta.
03:43Now, let's try to find out the velocity from these equations.
03:471. At any point above the axis, that is, where theta is equal to 0 degree, the velocity of the object will be around the axis.
03:57Because theta is equal to 0 degree is equal to cos theta 1.
04:022. At meru, that is, at theta is equal to 90 degree, the velocity of the object will remain unchanged.
04:10Because cos theta will be equal to 0.
04:133. If an object is taken from Meru in the direction of the Earth's equator, its gravitational force will remain at the axis.
04:22Because there, the force of the uncertain ball will be constant.
04:26In the next video, we will see Kepler's Theorem, the second part of the house equation.
04:32Keep watching.