请解析这个题目---**Question 7**
**Question Stem:**
There is a spherical cavity of radius R in a planet of radius 2R, as shown. The distance between center of two spheres d_OO' = R. The planet has a uniform mass density ρ (Except the cavity). The gravitational field at point P, which is a distance R to point O, is:
**Options:**
A. \frac{\rho\pi GR}{4}
B. \frac{\rho\pi GR}{2}
C. \rho\pi GR
D. 2\rho\pi GR
E. 4\rho\pi GR
**Chart/Diagram Description:**
* **Type:** Cross-sectional diagram showing a larger circle with a smaller circular cavity inside.
* **Main Elements:**
* **Large Circle:** Represents the planet. Its radius is labeled as 2R, indicated by an arrow extending from the center O to the outer edge.
* **Small Circle:** Represents the spherical cavity. Its radius is labeled as R, indicated by an arrow extending from the center O' to the edge of the small circle.
* **Centers:** O is the center of the large circle (planet). O' is the center of the small circle (cavity).
* **Distance between Centers:** The distance between O and O' is labeled as R, indicated by an arrow between O' and O.
* **Point P:** A point located along the line passing through O and O'. P is to the right of O, and the distance from O to P is labeled as R, indicated by an arrow between O and P.
* **Shaded Region:** The area between the outer circle and the inner circle is shaded, representing the planet's material with uniform mass density ρ. The density ρ is labeled within the shaded region.
* **Labels:** O, O', P, R, 2R, ρ.
* **Axis:** A horizontal line passes through O', O, and P, with distances R indicated between O' and O, and between O and P. An arrow points from O to P, indicating a distance of R. Another arrow points from O' to the edge of the cavity, indicating radius R. An arrow points from O to the outer edge of the planet, indicating radius 2R.
* **Relative Position:** O' is to the left of O. P is to the right of O. O', O, and P are collinear. The distance O'O is R, and the distance OP is R.
* **Annotations:** There is some handwritten text on the image, including "d_OO'", "ρ", "m", "m'", "P", "O", "O'", "R", "R", "2R", and some calculations like "m(-R) + 7m(x) = 0 / 8m" and "R/7". These handwritten annotations appear to be notes or scratch work, not part of the original problem statement or diagram labels.
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这是一个关于球体空腔引力场的物理问题。我们有一个半径为2R的星球,内部有一个半径为R的球形空腔。两个球心之间的距离为R。星球的密度为ρ。我们需要求出距离O点R处的P点的引力场强度。
解决这个问题的关键是使用叠加原理。我们可以将带空腔的星球看作一个完整的半径2R的球体减去一个半径R的"缺失"球体。这样,我们只需要分别计算完整球体和缺失球体在P点产生的引力场,然后将两者相减,就能得到带空腔星球在P点的引力场。
现在我们开始具体计算。首先计算完整球体在P点产生的引力场。由于P点距离球心O的距离为R,小于球体半径2R,所以P点位于球体内部。对于均匀球体内部的引力场,我们使用公式:E等于负的三分之四π G ρ r,其中r是从球心指向该点的位移矢量。在我们的情况下,r等于R,所以完整球体在P点的引力场为负的三分之四π G ρ R。
接下来计算缺失球体在P点产生的引力场。P点距离缺失球体的球心O'的距离为2R,大于缺失球体的半径R,所以P点位于缺失球体外部。对于球体外部的引力场,我们使用公式:E等于负的GM除以r的平方。缺失球体的质量为三分之四π R立方ρ,距离为2R。经过计算,缺失球体在P点的引力场为负的三分之一π G ρ R。
现在我们应用叠加原理得到最终结果。带空腔星球的引力场等于完整球体的引力场减去缺失球体的引力场。将前面计算的结果代入:负的三分之四π G ρ R减去负的三分之一π G ρ R,等于负π G ρ R。因此,引力场强度的模长为π G ρ R。这对应选项C,所以答案是C。