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M3 2026 Jan A Q2

A Level / Edexcel / M3

IAL 2026 Jan A Paper · Question 2

题目

Problem

2.

A spacecraft SS of mass mm moves in a straight line towards the centre of the Earth.

The Earth is modelled as a sphere of radius RR and SS is modelled as a particle.

When SS is at a distance xx, x⩾Rx\geqslant R, from the centre of the Earth, the force exerted by the Earth on SS is directed towards the centre of the Earth.

The force has magnitude Kx2\dfrac K{x^2}, where KK is a constant.

(a) Show that K=mgR2K=mgR^2

(2)

When SS is at a distance 3R3R from the centre of the Earth, the speed of SS is VV.

Assuming that air resistance can be ignored,

(b) find, in terms of gg, RR and VV, the speed of SS as it hits the surface of the Earth.

(7)
题目中文翻译

质量为 mm 的航天器 SS 沿直线向地心运动。

将地球视为半径为 RR 的球体，SS 视为质点。

SS 距地心为 xx、x⩾Rx\geqslant R 时，地球对 SS 施加的力指向地心。

该力的大小为 Kx2\dfrac K{x^2}，其中 KK 为常数。

(a) 证明 K=mgR2K=mgR^2。

(2)

SS 距地心为 3R3R 时的速率为 VV。

假设空气阻力可忽略，

(b) 用 gg、RR 和 VV 表示 SS 撞上地球表面时的速率。

(7)

解答

(a)

解法一

思路

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在地球表面，反平方引力必须等于物体的重量。注意题目中的 xx 是地心距离。

答题过程

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At the Earth’s surface x=Rx=R, and the gravitational force has magnitude mgmg. Thus

KR2=mg⟹K=mgR2.\frac K{R^2}=mg \quad\Longrightarrow\quad \boxed{K=mgR^2.}

(b)

解法一

思路

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飞船向内运动时，速率为正，但地心距离在减小，所以 dx/dt=−v\mathrm dx/\mathrm dt=-v。在链式法则中保留这个负号，再用初始条件确定积分常数。

答题过程

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Let vv denote the inward speed at distance xx from the Earth’s centre. Taking inward as positive for the acceleration,

dvdt=gR2x2,dxdt=−v.\frac{\mathrm dv}{\mathrm dt}=\frac{gR^2}{x^2}, \qquad \frac{\mathrm dx}{\mathrm dt}=-v.

Therefore

−vdvdx=gR2x2.-v\frac{\mathrm dv}{\mathrm dx}=\frac{gR^2}{x^2}.

Separating variables and integrating,

∫v dv=−gR2∫x−2 dx,12v2=gR2x+C.\int v\,\mathrm dv=-gR^2\int x^{-2}\,\mathrm dx, \qquad \frac12v^2=\frac{gR^2}{x}+C.

When x=3Rx=3R, v=Vv=V, so

C=12V2−gR3.C=\frac12V^2-\frac{gR}{3}.

At the surface, x=Rx=R. Hence

12v2=gR+12V2−gR3=12V2+2gR3.\frac12v^2=gR+\frac12V^2-\frac{gR}{3} =\frac12V^2+\frac{2gR}{3}.

Taking the positive root for a speed,

v=V2+4gR3.\boxed{v=\sqrt{V^2+\frac{4gR}{3}}.}

解法二

思路

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也可以对变力做功积分。若使用向外的位置坐标，引力为负，而积分上限小于下限，最终得到的功仍为正。

答题过程

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With outward distance xx as coordinate, the signed gravitational force is

Fx=−mgR2x2.F_x=-\frac{mgR^2}{x^2}.

The work done by gravity as SS moves from x=3Rx=3R to x=Rx=R is

W=∫3RR−mgR2x2 dx=[mgR2x]3RR=mgR−mgR3=2mgR3.W=\int_{3R}^{R}-\frac{mgR^2}{x^2}\,\mathrm dx =\left[\frac{mgR^2}{x}\right]_{3R}^{R} =mgR-\frac{mgR}{3}=\frac{2mgR}{3}.

By the work–energy principle,

12mv2−12mV2=2mgR3,\frac12mv^2-\frac12mV^2=\frac{2mgR}{3},

so the impact speed is

v=V2+4gR3.\boxed{v=\sqrt{V^2+\frac{4gR}{3}}.}