高一物理《三4人造卫星宇宙速度》课件.ppt

[例2]　高度不同的三颗人造卫星，某一瞬间的位置恰好与地心在同一条直线上，如图3－4－3所示，则此时它们的线速度大小、角速度大小、周期和向心加速度的大小比较为 (　　) A．ω1＞ω2＞ω3　　　　　　 B．v1＜v2＜v3 C．T1＝T2＝T3 D．a1＞a2＞a3 图3－4－3 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. [答案]　B Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 2．火星有两颗卫星，分别是火卫一和火卫二，它们的轨 道近似为圆，已知火卫一的周期为7小时39分，火卫二的周期为30小时18分，则两颗卫星相比 (　　) A．火卫一距火星表面较近 B．火卫二的角速度较大 C．火卫一的运行速度较大 D．火卫二的向心加速度较大 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 返回 理解教材新知 把握热点考向 应用创新演练 第4节 知识点一 知识点二 考向二 第三章 考向一 随堂基础巩固 课时跟踪训练 考向三 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. 1.人造地球卫星绕地球做匀速圆 周运动的向心力是由地球对卫 星的万有引力提供，其轨道的 圆心与地心重合。 2．人造地球卫星在不同的轨道 上线速度不同，第一宇宙速度v＝7.9 km/s是卫星绕 地球做圆周运动的最大环绕速度。 3．同步卫星的线速度、角速度、周期、轨道、向心加 速度均是一定的。 Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004-2011 Aspose Pty Ltd. Evaluation only. Created with Aspose.Slides for .NET 3.5 Client Profile 5.2.0.0. Copyright 2004