沖量矩與角動(dòng)量(1)(已修改)
2025-05-28 22:21 本頁面
【正文】 沖量矩與角動(dòng)量 1 Johannes Kepler (1571 1630) Plaary orbits are ellipses with the sun at one focus. Figure shows the shape and features of an ellipse. Plaary orbits are not as eccentric as the one drawn here, however. Kepler’s first law 地球軌道的偏心率為 ? Kepler’s Laws 引言 沖量矩與角動(dòng)量 2 The line joining a pla to the sun sweeps out equal amounts of area in equal amounts of time. Figure 2 shows this area. If the time interval is equal on both sides, the green area will equal the blue area. mLvrdtdS221 0??? ?? 開普勒第二定律 Kepler’s second law 沖量矩與角動(dòng)量 3 The period of the orbit to the second power is equal to the semimajor axis to the third. The period must be in years and the semimajor axis in astronomical units for this formula to work Kepler’s third law eGmaT 232 4 ??Kepler39。s second and third laws can be stated in ordinary language as: ?Plas move faster near perihelion than near aphelion ?Plas that are close to the sun orbit faster than more distant plas. 沖量矩與角動(dòng)量 4 開普勒定律 行星沿橢圓軌道繞太陽運(yùn)行，太陽位于橢圓的一個(gè)焦點(diǎn)上。行星軌道的偏心率都比較小，很接近圓。 對(duì)任一行星，它的位置矢量（以太陽中心為參考點(diǎn)）在相等的時(shí)間內(nèi)掃過相等的面積。 —— 面積定律 恒量?32aT行星繞太陽運(yùn)動(dòng)周期 T的平方和橢圓軌道的半長(zhǎng)軸 a的立方成正比，即： mLvrdtdS221 0??? ??eGmaT 232 4 ?? 沖量矩與角動(dòng)量 5 ? O 勻速直線運(yùn)動(dòng) 面積定理成立??！ 動(dòng)量和動(dòng)能均守恒 動(dòng)量和動(dòng)能均發(fā)生變化 動(dòng)量和動(dòng)能都不是對(duì)上述現(xiàn)象做出統(tǒng)一描述的物理量！ vrdtdS ?? ?? 21問題的提出： 沖量矩與角動(dòng)量 6 質(zhì)點(diǎn)的角動(dòng)量 角動(dòng)量守恒定律 O ?平均角速度 Average angular velocity ttttav ????? )()( ???瞬時(shí)角速度 Instantaneous angular velocity ??? ?dtd??It is a Vector Direction: righthand rule ??Unit: radians per second 沖量矩與角動(dòng)量 7 ?角加速度 Angular Acceleration ???? ?22dtddtd ????Constant angular acceleration t??? ?? 022dtrddtvda ??? ??22100)( ttt ???? ???)(2 0202 ????? ??? )(2 0202221000xxavvattvxxatvv????????Linear motion expression ?? rarv t ?? , 沖量矩與角動(dòng)量 8 ?質(zhì)點(diǎn)的角動(dòng)量 Angular momentum of a point mass vmp ?? ??s i nr m vL ?Magnitude ： r?vm??O Linear momentum ,0??iiP?O m m r r v?v?rm
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