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题型分析 — Gravitational Fields

Question Type 1: 引力场强度计算

如何识别

给出天体质量 MM 和半径 RR,求表面或某高度的引力场强度 gg

标准解题方法

方法
  1. 确定目标点到中心距离 rr
  2. g=GM/r2g = GM/r^2
  3. 若求百分比变化:计算新旧 gg 之比

评分标准

评分要点
  • C1: g=GM/r2g = GM/r^2
  • C1: 正确代入 GG, MM, rr
  • A1: 答案含单位

完整原题

Example 1 — 9702_w20_qp_41 (6 marks):

An isolated planet may be assumed to be a uniform sphere of radius 3.39×1063.39 \times 10^6 m with its mass of 6.42×10236.42 \times 10^{23} kg concentrated at its centre. (b) Calculate the gravitational field strength at the surface of the planet. (c) Calculate the height above the surface of the planet at which the gravitational field strength is 1.0% less than its value at the surface.

Details

📝 MS 展开查看 MS:

  • C1: g=GM/R2g = GM/R^2
  • C1: =(6.67×1011×6.42×1023)/(3.39×106)2= (6.67 \times 10^{-11} \times 6.42 \times 10^{23})/(3.39 \times 10^6)^2
  • A1: g=3.73g = 3.73 N kg1^{-1}
  • C1: 0.99×3.73=GM/r20.99 \times 3.73 = GM/r^2
  • C1: r=3.41×106r = 3.41 \times 10^6 m
  • A1: height =rR=2×104= r - R = 2 \times 10^4 m

Example 2 — 9702_w23_qp_41 (4 marks):

Fig. 1.2 shows the variation with xx of the gravitational field gg at point PP due to a sphere. The magnitude of the gravitational field at the surface of the sphere is YY. Calculate the field at 2R2R from centre.

Details

📝 MS 展开查看 MS:

  • C1: g1/r2g \propto 1/r^2
  • C1: field at 2R=Y/42R = Y/4
  • A1: Y/4Y/4

常见陷阱

注意
  • rr中心算起,不是表面
  • 区分 GGgg
  • 百分比减少是 0.99g0.99g 不是 0.01g0.01g

Question Type 2: 卫星轨道与开普勒第三定律

如何识别

给轨道周期/半径求半径/周期,或证明 T2r3T^2 \propto r^3

标准解题方法

推导步骤
  1. 引力 = 向心力:GMm/r2=mrω2GMm/r^2 = mr\omega^2
  2. ω=2π/T\omega = 2\pi/T
  3. GM/r3=4π2/T2GM/r^3 = 4\pi^2/T^2
  4. T2=(4π2/GM)r3T^2 = (4\pi^2/GM)r^3

评分标准

评分
  • B1: gravitational force provides centripetal force
  • M1: 写出方程
  • A1: 完成推导

完整原题

Example 1 — 9702_s21_qp_41 (6 marks):

A satellite of mass 1200 kg is in a circular orbit about the Earth. The period of the orbit is 94 minutes. (b) Calculate the radius of the orbit. (c) The satellite enters a new orbit with period 150 minutes. (i) Show that the radius of the new orbit is 9.4×1069.4 \times 10^6 m. (ii) State whether the gravitational potential energy increases or decreases. (iii) Determine the magnitude of the change in gravitational potential energy.

Details

📝 MS 展开查看 MS:

  • C1: GMm/r2=mrω2GMm/r^2 = mr\omega^2 and ω=2π/T\omega = 2\pi/T
  • C1: 6.67×1011×6.0×1024=r3×[2π/(94×60)]26.67 \times 10^{-11} \times 6.0 \times 10^{24} = r^3 \times [2\pi/(94 \times 60)]^2
  • A1: r=6.9×106r = 6.9 \times 10^6 m
  • C1: r3/T2=constantr^3/T^2 = \text{constant}
  • A1: r3=(6.9×106)3×(150/94)2r^3 = (6.9 \times 10^6)^3 \times (150/94)^2, r=9.4×106r = 9.4 \times 10^6 m
  • B1: separation increases, so potential energy increases
  • C1: ΔEP=GMm(1/r11/r2)\Delta E_P = GMm(1/r_1 - 1/r_2)
  • C1: =6.67×1011×6.0×1024×1200×(1/6.9×1061/9.4×106)= 6.67 \times 10^{-11} \times 6.0 \times 10^{24} \times 1200 \times (1/6.9\times10^6 - 1/9.4\times10^6)
  • A1: =1.9×1010= 1.9 \times 10^{10} J

Example 2 — 9702_w22_qp_41 (7 marks):

(a) State the equation for the gravitational force between two point masses. (b) Show that T2=kR3T^2 = kR^3 for a satellite in circular orbit. (c) Calculate the radius of a geostationary orbit (T = 24 h, M = 6.0×10246.0 \times 10^{24} kg). (ii) State the other two conditions for a geostationary orbit.

Details

📝 MS 展开查看 MS:

  • M1: F=Gm1m2/r2F = Gm_1m_2/r^2
  • A1: identifies GG as gravitational constant
  • B1: gravitational force provides centripetal force
  • M1: mRω2=GMm/R2mR\omega^2 = GMm/R^2 and ω=2π/T\omega = 2\pi/T
  • A1: completes algebra to T2=(4π2/GM)R3T^2 = (4\pi^2/GM)R^3
  • C1: (24×3600)2=(4π2×R3)/(6.67×1011×6.0×1024)(24 \times 3600)^2 = (4\pi^2 \times R^3)/(6.67 \times 10^{-11} \times 6.0 \times 10^{24})
  • A1: R=4.2×107R = 4.2 \times 10^7 m
  • B1: above the Equator
  • B1: west to east

常见陷阱

注意
  • 卫星质量 mm 在推导中消去
  • 周期必须用秒(s)
  • geostationary 要三个条件:24 h、赤道上空、西向东

Question Type 3: 引力势与引力势能

如何识别

求引力势 ϕ\phi 或势能 EPE_P,或计算逃逸能量。

标准解题方法

定义题

Gravitational potential at a point = work done per unit mass in bringing a small test mass from infinity to that point.

ϕ=GM/r\phi = -GM/r,越近越负。

引力势能 EP=mϕ=GMm/rE_P = m\phi = -GMm/r

评分标准

评分要点
  • M1: work done per unit mass
  • A1: from infinity
  • C1: ϕ=GM/r\phi = -GM/r
  • A1: 数字正确

完整原题

Example 1 — 9702_s24_qp_41 (5 marks):

(a) Define gravitational potential at a point. (b) Satellite X of mass M orbits at distance 4R, satellite Y of mass 2M orbits at distance R. Gravitational potential at X is ϕ-\phi. (ii) State the gravitational potential at Y in terms of ϕ\phi.

Details

📝 MS 展开查看 MS:

  • B1: work done per unit mass
  • B1: (bringing small test mass) from infinity (to the point)
  • B1: ϕ1/r\phi \propto 1/r
  • B1: at Y, r=Rr = R, which is 1/4 of distance to X
  • B1: so potential at Y = 44 times at X = 4ϕ-4\phi

Example 2 — 9702_w23_qp_42 (4 marks):

Define gravitational potential. Calculate gravitational potential at the surface of the Moon. Mass of Moon = 7.35×10227.35 \times 10^{22} kg, radius = 1.74×1061.74 \times 10^6 m.

Details

📝 MS 展开查看 MS:

  • B1: work done per unit mass
  • B1: from infinity (to the point)
  • C1: ϕ=GM/r\phi = -GM/r
  • A1: ϕ=(6.67×1011×7.35×1022)/(1.74×106)\phi = -(6.67 \times 10^{-11} \times 7.35 \times 10^{22})/(1.74 \times 10^6)
  • A1: ϕ=2.82×106\phi = -2.82 \times 10^6 J kg1^{-1}

常见陷阱

注意
  • 势能是负的!负号不能丢
  • "from infinity" 是定义关键
  • 势能变化:ΔEP=EP2EP1\Delta E_P = E_{P2} - E_{P1}
  • 逃逸时 12mv2=GMm/r\frac12 mv^2 = GMm/r