爭議題 – Gauss’s Law of Gravity & Mass Distribution

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爭議題 – 高斯定理之引力場強 Gauss’s Law of Gravity

清華大學學士後醫學院 112 年度物理試題 46 吳祖諺 筆

The mass density inside a planet is generally not uniform. However, the mass density of the concentric shells of a spherical planet is approximately the same when the planet is decomposed into concentric spherical shells. As a result, for a spherical planet �� of radius ��, the mass density can be treated as a function of its distance to the center of the planet

��(��)=��0 (1 �� 2��)

Assuming the gravitational field strength at the planet’s surface is ����>�� ≝��0. Show that at �� = 1 2 ��, the gravitational field has a magnitude smaller than half that of planet ��'s surface but is not zero, i.e., ����>½�� < 1 2��0 and ����>½�� ≠0

星球的整體內部一般擁有非均勻的質量密度,然而若是將球體分成多重同心球殼的話, 每個同心球殼內的質量密度皆大略相同。假設有顆星球是一個半徑為 �� 、質量為 ���� 的非 均質球體,吾人稱之為星球 �� 。其質量密度為 ��(��)=��0 (1 �� 2��) 且 ��0 為常數,�� 為距離

星球 �� 中心位置之距。試證明當同心球體的半徑變為星球 �� 的半徑之一半時,其重力場 強度會比在星球 �� 表面的重力場強度之一半弱,即 ����>½�� <½��0 且 ����>½�� ≠0。

Analysis:

A planet’s density usually increases as we go inwards from the crust to core Suppose that a planet called �� is a sphere of radius �� and that its density �� is modeled by ��(��)=��0 (1 �� 2��)

Where �� is the distance from the center, ��0 and �� are constants. The approximate volume of a spherical shell of radius �� and thickness ���� is given by ���� =4����2���� so the mass in this shell is given by ���� =��(��)���� =4����0 (1 �� 2��)��2 ���� Therefore, the total mass in general is given by the integral:

題意:

For the total mass of the planet �� with radius �� =��, we have:

For the total mass of the inner spherical shell with radius ��

, we have:

Note that: 積分極坐標變換 ∯���� �� = ∬��2������ ���� ���������� ���� =∫������ =4��∫��0 (1 �� 2��)��2���� �� 0 �� 0 =4����0[��3(8�� 3��) 24�� |0 ��]=����0 (8�� 3��) 6�� ��3
���� =����0 ��3(8�� 3��) 6�� =����0 5 6 ��3
1 2 ��
��½�� =����0 (��2)3[8�� 3(��2)] 6�� =����0 13 96 ��3 ����:��½�� =(56)(9613)≈615 Given Gauss’s law: ∯��⋅��̂���� �� = 4������������ ⇒∯|��||��̂|cos(��)���� �� = ��∯���� �� = ��(4����2) Therefore, we have ��(4����2)= 4������������ For �� >��: (星球��表面的場強:成反比) ����>�� ⋅4����2 = 4�������� ����>�� 4����2 = 4����(����0 5 6 ��3) ����>�� ≝��0 = 5 6��(����0��3 1 ��2) ⇒ 1 2��0 =0416⋅��(����0��3 1 ��2) For �� <��:(星球��內部的場強:成正比) ����<�� ⋅4����2 = 4�������� ����<�� ⋅4����2 = 4����[����0 (8�� 3��) 6�� ��3] ����<�� = 1 6��[����0 (8�� 3��) �� ��]=0.16⋅��(����0 8�� 3�� �� ��) Note that: ⌘ ��⋅��̂����=��⋅���� ⌘ �� is the radius of Gaussian surface �� ⌘ �� is the radius of planet �� ⌘ |��̂|=1 ⌘ cos(��)= 1 The ratio of ���� &��½�� 差約六倍 ���� 之 構 成 元 素
=
For �� >½��: (星球��內部之同心球的表面場強:成反比) ����>½�� 4����2 = 4������½�� ����>½�� ⋅4����2 = 4����(����0 13 96 ��3) ����>½�� = 13 96��(����0��3 1 ��2)=0.135416 ��(����0��3 1 ��2) Given the fact that we have {����>½�� =0135416⋅��(����0��3 1 ��2) ½����>�� =0416⋅��(����0��3 1 ��2) By comparing 1 2��0 and ����>½�� we have ½��0:����>½�� =(512)÷(1396)=40 13 =3.076923 This shows that ����>½�� < ½��0 and ����>½�� ≠0 …………………… 畢證 星球��內之同心球的表面場強 星球��表面 外 場強的一半 兩者場強差約三倍

The original question is presented as follows:

With a hint provided as follows:

*The above says that ��������(��) is at least 4 times of ��������

and hence

��
)
��(
2
1 2
(
2
��
)<
��(��)

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