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dsProject/dsLightRag/static/YunXiao/《万有引力定律》试题【markdown】/简单_3.md

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> 1\.
> 假设在地球周围有质量相等的A、B两颗地球卫星已知地球半径为R卫星A距地面高度为R卫星B距地面高度为2R卫星B受到地球的万有引力大小为F
> 卫星A受到地球的万有引力大小为
>
> A
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> ![](https://dsideal.obs.cn-north-1.myhuaweicloud.com/HuangHai/WToM/images/467fa785b554fd88ee0f7ce1e71d8ed4/media/image1.png)
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> B
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> ![](https://dsideal.obs.cn-north-1.myhuaweicloud.com/HuangHai/WToM/images/467fa785b554fd88ee0f7ce1e71d8ed4/media/image2.png)
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>
>
> C
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> ![](https://dsideal.obs.cn-north-1.myhuaweicloud.com/HuangHai/WToM/images/467fa785b554fd88ee0f7ce1e71d8ed4/media/image3.png)
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> D 4F
>
> 【难度】\
> 简单
>
> 【答案】
>
> C
>
> 【解析】
>
> 【解答】B卫星距地心为3R根据万有引力的表达式可知受到的万有引力为
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> ![](https://dsideal.obs.cn-north-1.myhuaweicloud.com/HuangHai/WToM/images/467fa785b554fd88ee0f7ce1e71d8ed4/media/image4.png)
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> A卫星距地心为2R受到的万有引力为
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> ![](https://dsideal.obs.cn-north-1.myhuaweicloud.com/HuangHai/WToM/images/467fa785b554fd88ee0f7ce1e71d8ed4/media/image5.png)
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> ,则有
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> ![](https://dsideal.obs.cn-north-1.myhuaweicloud.com/HuangHai/WToM/images/467fa785b554fd88ee0f7ce1e71d8ed4/media/image6.png)
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> ABD不符合题意C符合题意.
>
> 故答案为C
>
> 【分析】卫星做圆周运动,万有引力提供向心力,结合卫星的轨道半径,卫星与地球之间的万有引力可以利用公式来计算比较即可。
>
> 【关联章节】第七章 万有引力与宇宙航行第七章 万有引力与宇宙航行 \> 2.
> 万有引力定律;
>
> 【关联知识点】经典力学的成就与局限性 \> 万有引力定律;
>
> 【关联素养点】暂无数据
>
> 【关联关键能力】暂无数据