Companion to the IB-Style Practice SetIB 风格练习题的解析配套
Syllabus D1.1 to D1.6考纲 D1.1 至 D1.6PHYSICS HL
Two identical spheres of mass $8.0\times 10^{3}\ \mathrm{kg}$, centres $5.0\ \mathrm{m}$ apart. (a) the attractive force; (b) the new force when the separation is tripled, with reasoning.两个质量 $8.0\times 10^{3}\ \mathrm{kg}$ 的相同球体，中心相距 $5.0\ \mathrm{m}$。(a) 引力；(b) 间距增至三倍后的新引力及理由。
Both masses equal, $M = m = 8.0\times 10^{3}\ \mathrm{kg}$, $r = 5.0\ \mathrm{m}$. Use Newton's law of gravitation $F = \dfrac{GMm}{r^{2}}$. (M1)
$$ F = \frac{(6.67\times 10^{-11})(8.0\times 10^{3})^{2}}{(5.0)^{2}} = \frac{(6.67\times 10^{-11})(6.4\times 10^{7})}{25} \approx 1.7\times 10^{-4}\ \mathrm{N}. $$(A1)
The force is inverse-square in $r$, so tripling $r$ divides $F$ by $3^{2} = 9$. (R1)
$$ F' = \frac{F}{9} = \frac{1.71\times 10^{-4}}{9} \approx 1.9\times 10^{-5}\ \mathrm{N}. $$(A1)
两质量相等，$M = m = 8.0\times 10^{3}\ \mathrm{kg}$，$r = 5.0\ \mathrm{m}$。用万有引力定律 $F = \dfrac{GMm}{r^{2}}$。(M1)
$$ F = \frac{(6.67\times 10^{-11})(8.0\times 10^{3})^{2}}{(5.0)^{2}} = \frac{(6.67\times 10^{-11})(6.4\times 10^{7})}{25} \approx 1.7\times 10^{-4}\ \mathrm{N}. $$(A1)
引力对 $r$ 为反平方，故 $r$ 增至三倍使 $F$ 除以 $3^{2} = 9$。(R1)
$$ F' = \frac{F}{9} = \frac{1.71\times 10^{-4}}{9} \approx 1.9\times 10^{-5}\ \mathrm{N}. $$(A1)
Mars, $M = 6.4\times 10^{23}\ \mathrm{kg}$, $R = 3.4\times 10^{6}\ \mathrm{m}$. (a) surface field strength; (b) field strength at $r = 2R$, with justification.火星，$M = 6.4\times 10^{23}\ \mathrm{kg}$，$R = 3.4\times 10^{6}\ \mathrm{m}$。(a) 表面场强度；(b) $r = 2R$ 处的场强度及依据。
Field strength is the force per unit mass, $g = \dfrac{GM}{r^{2}}$, evaluated at $r = R$. (M1)
$$ g = \frac{(6.67\times 10^{-11})(6.4\times 10^{23})}{(3.4\times 10^{6})^{2}} \approx 3.7\ \mathrm{N\,kg^{-1}}. $$(A1)
Field strength is inverse-square in $r$; doubling $r$ to $2R$ divides $g$ by $2^{2} = 4$. (R1)
$$ g' = \frac{g}{4} = \frac{3.69}{4} \approx 0.92\ \mathrm{N\,kg^{-1}}. $$(A1)
场强度为单位质量受力，$g = \dfrac{GM}{r^{2}}$，取 $r = R$。(M1)
$$ g = \frac{(6.67\times 10^{-11})(6.4\times 10^{23})}{(3.4\times 10^{6})^{2}} \approx 3.7\ \mathrm{N\,kg^{-1}}. $$(A1)
场强度对 $r$ 为反平方；$r$ 加倍到 $2R$ 使 $g$ 除以 $2^{2} = 4$。(R1)
$$ g' = \frac{g}{4} = \frac{3.69}{4} \approx 0.92\ \mathrm{N\,kg^{-1}}. $$(A1)
Satellite at $r = 8.0\times 10^{6}\ \mathrm{m}$, planet $M = 6.0\times 10^{24}\ \mathrm{kg}$. (a) show $v = \sqrt{GM/r}$; (b) orbital speed and period; (c) Kepler factor for a second satellite at four times the radius.卫星位于 $r = 8.0\times 10^{6}\ \mathrm{m}$，行星 $M = 6.0\times 10^{24}\ \mathrm{kg}$。(a) 证明 $v = \sqrt{GM/r}$；(b) 轨道速度与周期；(c) 四倍半径处第二颗卫星的开普勒倍率。
Gravity provides the centripetal force for the circular orbit, so set $\dfrac{GMm}{r^{2}} = \dfrac{mv^{2}}{r}$. (M1)
The orbiting mass $m$ cancels and one power of $r$ cancels, leaving $v^{2} = \dfrac{GM}{r}$, hence $v = \sqrt{\dfrac{GM}{r}}$. (A1)
(M1·A1)
The period is the circumference divided by the speed, $T = \dfrac{2\pi r}{v} = \dfrac{2\pi (8.0\times 10^{6})}{7.07\times 10^{3}} \approx 7.1\times 10^{3}\ \mathrm{s}$. (A1)
$T^{2}\propto r^{3}$ gives $T\propto r^{3/2}$. With the radius multiplied by $4$, $T$ multiplies by $4^{3/2} = 8$. (B1)
引力为圆轨道提供向心力，故令 $\dfrac{GMm}{r^{2}} = \dfrac{mv^{2}}{r}$。(M1)
绕行质量 $m$ 约去，一次 $r$ 约去，得 $v^{2} = \dfrac{GM}{r}$，故 $v = \sqrt{\dfrac{GM}{r}}$。(A1)
(M1·A1)
周期为周长除以速度，$T = \dfrac{2\pi r}{v} = \dfrac{2\pi (8.0\times 10^{6})}{7.07\times 10^{3}} \approx 7.1\times 10^{3}\ \mathrm{s}$。(A1)
$T^{2}\propto r^{3}$ 给出 $T\propto r^{3/2}$。半径乘以 $4$，$T$ 乘以 $4^{3/2} = 8$。(B1)
Planet $M = 6.0\times 10^{24}\ \mathrm{kg}$, $R = 6.4\times 10^{6}\ \mathrm{m}$; probe $1500\ \mathrm{kg}$ on the surface. (a) surface potential and why it is negative; (b) potential energy of the probe; (c) define an equipotential and the work to move along it.行星 $M = 6.0\times 10^{24}\ \mathrm{kg}$，$R = 6.4\times 10^{6}\ \mathrm{m}$；探测器 $1500\ \mathrm{kg}$ 在表面。(a) 表面势及其为负的原因；(b) 探测器的势能；(c) 定义等势面及沿其移动所做的功。
Use $V_g = -\dfrac{GM}{r}$ with $r = R$. (M1)
$$ V_g = -\frac{(6.67\times 10^{-11})(6.0\times 10^{24})}{6.4\times 10^{6}} \approx -6.3\times 10^{7}\ \mathrm{J\,kg^{-1}}. $$(A1)
It is negative because the zero of potential is chosen at infinity; bringing a mass from infinity to the surface, gravity does positive work, so the potential ends up below zero. (R1)
Potential energy is mass times potential, $E_p = m V_g$ (equivalently $E_p = -GMm/r$). (M1)
$$ E_p = (1500)(-6.25\times 10^{7}) \approx -9.4\times 10^{10}\ \mathrm{J}. $$(A1)
An equipotential surface is a surface on which the gravitational potential $V_g$ has the same value at every point; around a point mass these are concentric spheres. (A1)
The work done is $W = m\,\Delta V_g$. Along one equipotential $\Delta V_g = 0$, so $W = 0$. (A1)
This is consistent with the field being perpendicular to the equipotential: no component of the gravitational force acts along the path. (R1)
用 $V_g = -\dfrac{GM}{r}$ 取 $r = R$。(M1)
$$ V_g = -\frac{(6.67\times 10^{-11})(6.0\times 10^{24})}{6.4\times 10^{6}} \approx -6.3\times 10^{7}\ \mathrm{J\,kg^{-1}}. $$(A1)
它为负，因为势的零点取在无穷远；把质量从无穷远移到表面时引力做正功，故势最终低于零。(R1)
势能为质量乘以势，$E_p = m V_g$（等价于 $E_p = -GMm/r$）。(M1)
$$ E_p = (1500)(-6.25\times 10^{7}) \approx -9.4\times 10^{10}\ \mathrm{J}. $$(A1)
等势面是其上每点引力势 $V_g$ 取同一值的曲面；点质量周围为同心球面。(A1)
做功为 $W = m\,\Delta V_g$。沿同一等势面 $\Delta V_g = 0$，故 $W = 0$。(A1)
这与场垂直于等势面一致：引力沿路径无分量。(R1)
Planet $M = 6.0\times 10^{24}\ \mathrm{kg}$, $R = 6.4\times 10^{6}\ \mathrm{m}$. (a) derive $v_{esc} = \sqrt{2GM/r}$ from zero total energy; (b) calculate the surface escape speed; (c) why it is independent of the escaping mass, and that $v_{esc} = \sqrt{2}\,v_{orbit}$.行星 $M = 6.0\times 10^{24}\ \mathrm{kg}$，$R = 6.4\times 10^{6}\ \mathrm{m}$。(a) 由总能量为零推导 $v_{esc} = \sqrt{2GM/r}$；(b) 计算表面逃逸速度；(c) 它为何与逃逸质量无关，以及 $v_{esc} = \sqrt{2}\,v_{orbit}$。
"Just escaping" means reaching $r \to \infty$ with $v \to 0$, where both $E_k$ and $E_p$ vanish, so the total energy is zero. By conservation, at launch the total energy is also zero: (M1)
$$ \tfrac{1}{2} m v_{esc}^{2} - \frac{GMm}{r} = 0. $$(M1 for setting up the energy balance)
The mass $m$ cancels and rearranging gives $v_{esc} = \sqrt{\dfrac{2GM}{r}}$. (A1)
(M1·A1)
The escaping mass $m$ multiplies both $E_k$ and $E_p$, so it cancels from the energy equation: escape speed depends only on $G$, the planet mass $M$ and the distance $r$. (R1)
Compare with the circular orbital speed $v_{orbit} = \sqrt{GM/r}$. (M1)
$$ \frac{v_{esc}}{v_{orbit}} = \frac{\sqrt{2GM/r}}{\sqrt{GM/r}} = \sqrt{2}, \qquad \text{so } v_{esc} = \sqrt{2}\,v_{orbit}. $$(A1)
"恰好逃逸"指到达 $r \to \infty$ 时 $v \to 0$，此处 $E_k$ 与 $E_p$ 都为零，故总能量为零。由守恒，在发射时总能量也为零：(M1)
$$ \tfrac{1}{2} m v_{esc}^{2} - \frac{GMm}{r} = 0. $$(列出能量平衡得 M1)
质量 $m$ 约去，整理得 $v_{esc} = \sqrt{\dfrac{2GM}{r}}$。(A1)
(M1·A1)
逃逸质量 $m$ 同时乘 $E_k$ 与 $E_p$，故从能量方程中约去：逃逸速度只取决于 $G$、行星质量 $M$ 与距离 $r$。(R1)
与圆轨道速度 $v_{orbit} = \sqrt{GM/r}$ 比较。(M1)
$$ \frac{v_{esc}}{v_{orbit}} = \frac{\sqrt{2GM/r}}{\sqrt{GM/r}} = \sqrt{2}, \qquad \text{故 } v_{esc} = \sqrt{2}\,v_{orbit}. $$(A1)
Four moons; $\lg r$ and $\lg T$ tabulated. (a) show $\lg T$ vs $\lg r$ is linear and state the expected gradient; (b) calculate the gradient and compare with Kepler; (c) use the intercept $\lg c = -6.50$ to find $M$; (d) one advantage of the log graph.四颗卫星；列出 $\lg r$ 与 $\lg T$。(a) 证明 $\lg T$ 对 $\lg r$ 为直线并写出预期斜率；(b) 计算斜率并与开普勒比较；(c) 用截距 $\lg c = -6.50$ 求 $M$；(d) 对数图的一个优势。
Take logarithms (base 10) of $T^{2} = \dfrac{4\pi^{2}}{GM}r^{3}$, that is $T = \left(\dfrac{4\pi^{2}}{GM}\right)^{1/2} r^{3/2}$. (M1)
$$ \lg T = \tfrac{3}{2}\lg r + \lg\!\left[\left(\tfrac{4\pi^{2}}{GM}\right)^{1/2}\right]. $$This has the straight-line form $\lg T = m\,\lg r + \lg c$ with constant intercept, so the plot is linear. (A1)
The expected gradient is $\tfrac{3}{2} = 1.5$. (A1)
Use the two end points $(6.90,\,3.85)$ and $(7.60,\,4.90)$, which are widely separated: (M1)
$$ \text{gradient} = \frac{4.90 - 3.85}{7.60 - 6.90} = \frac{1.05}{0.70} = 1.5. $$(A1)
This equals the predicted $1.5$, so the data are consistent with Kepler's third law $T^{2}\propto r^{3}$. (A1)
The intercept satisfies $\lg c = \lg\left[\left(\tfrac{4\pi^{2}}{GM}\right)^{1/2}\right] = -6.50$, so $c = 10^{-6.50} = \left(\tfrac{4\pi^{2}}{GM}\right)^{1/2}$. (M1)
Square both sides: $c^{2} = 10^{-13.0} = \dfrac{4\pi^{2}}{GM}$. (M1)
$$ M = \frac{4\pi^{2}}{G\,c^{2}} = \frac{4\pi^{2}}{(6.67\times 10^{-11})(10^{-13.0})}. $$(A1)
$$ M = \frac{39.48}{6.67\times 10^{-24}} \approx 6.0\times 10^{24}\ \mathrm{kg}. $$(A1)
The radii and periods span more than an order of magnitude; a log-log plot compresses this wide range onto one readable straight line. (B1)
The gradient directly tests the power-law exponent ($1.5$), so the relationship $T\propto r^{3/2}$ is confirmed without first assuming it to plot $T^{2}$ against $r^{3}$. (R1)
对 $T^{2} = \dfrac{4\pi^{2}}{GM}r^{3}$（即 $T = \left(\dfrac{4\pi^{2}}{GM}\right)^{1/2} r^{3/2}$）取以 10 为底的对数。(M1)
$$ \lg T = \tfrac{3}{2}\lg r + \lg\!\left[\left(\tfrac{4\pi^{2}}{GM}\right)^{1/2}\right]. $$此式具有直线形式 $\lg T = m\,\lg r + \lg c$，截距为常量，故图为直线。(A1)
预期斜率为 $\tfrac{3}{2} = 1.5$。(A1)
用相距较远的两端点 $(6.90,\,3.85)$ 与 $(7.60,\,4.90)$：(M1)
$$ \text{斜率} = \frac{4.90 - 3.85}{7.60 - 6.90} = \frac{1.05}{0.70} = 1.5. $$(A1)
它等于预测值 $1.5$，故数据与开普勒第三定律 $T^{2}\propto r^{3}$ 相符。(A1)
截距满足 $\lg c = \lg\left[\left(\tfrac{4\pi^{2}}{GM}\right)^{1/2}\right] = -6.50$，故 $c = 10^{-6.50} = \left(\tfrac{4\pi^{2}}{GM}\right)^{1/2}$。(M1)
两边平方：$c^{2} = 10^{-13.0} = \dfrac{4\pi^{2}}{GM}$。(M1)
$$ M = \frac{4\pi^{2}}{G\,c^{2}} = \frac{4\pi^{2}}{(6.67\times 10^{-11})(10^{-13.0})}. $$(A1)
$$ M = \frac{39.48}{6.67\times 10^{-24}} \approx 6.0\times 10^{24}\ \mathrm{kg}. $$(A1)
半径与周期跨越一个数量级以上；双对数图把这一宽范围压缩到一条可读的直线上。(B1)
斜率直接检验幂律指数（$1.5$），故无需先假设关系即可确认 $T\propto r^{3/2}$，而不必直接作 $T^{2}$ 对 $r^{3}$ 图。(R1)
Probe measures $g$ against $1/r^{2}$ near a planet. (a) why $g$ vs $1/r^{2}$ is a line through the origin and what the gradient is; (b) gradient and planet mass $M$; (c) the relation between $g$ and $V_g$ and which graph feature gives $g$.探测器测量行星附近 $g$ 对 $1/r^{2}$。(a) 为何 $g$ 对 $1/r^{2}$ 为过原点直线及斜率含义；(b) 斜率与行星质量 $M$；(c) $g$ 与 $V_g$ 的关系及哪个图特征给出 $g$。
Field strength is $g = \dfrac{GM}{r^{2}} = (GM)\left(\dfrac{1}{r^{2}}\right)$. (M1)
This has the form $y = (\text{gradient})\,x$ with $x = 1/r^{2}$ and no intercept, so a graph of $g$ against $1/r^{2}$ is a straight line through the origin. (A1)
The gradient is $GM$. (A1)
Read the gradient from two widely separated points, e.g. $(1.60\times 10^{-15},\,0.64)$ and $(10.0\times 10^{-15},\,4.00)$: (M1)
$$ \text{gradient} = \frac{4.00 - 0.64}{(10.0 - 1.60)\times 10^{-15}} = \frac{3.36}{8.40\times 10^{-15}} \approx 4.0\times 10^{14}. $$(A1)
Since the gradient equals $GM$: $M = \dfrac{\text{gradient}}{G}$. (M1)
$$ M = \frac{4.0\times 10^{14}}{6.67\times 10^{-11}} \approx 6.0\times 10^{24}\ \mathrm{kg}. $$(A1)
The field strength is the negative gradient of the potential, $g = -\dfrac{\Delta V_g}{\Delta r}$. (A1)
So on a graph of $V_g$ against $r$, the field strength at a point equals minus the slope (the local gradient) of the curve. (A1)
The minus sign makes $g$ point toward decreasing potential, that is, inward toward the planet. (R1)
场强度为 $g = \dfrac{GM}{r^{2}} = (GM)\left(\dfrac{1}{r^{2}}\right)$。(M1)
此式具有 $y = (\text{斜率})\,x$ 形式，$x = 1/r^{2}$，无截距，故 $g$ 对 $1/r^{2}$ 的图为过原点的直线。(A1)
斜率为 $GM$。(A1)
用相距较远的两点读斜率，如 $(1.60\times 10^{-15},\,0.64)$ 与 $(10.0\times 10^{-15},\,4.00)$：(M1)
$$ \text{斜率} = \frac{4.00 - 0.64}{(10.0 - 1.60)\times 10^{-15}} = \frac{3.36}{8.40\times 10^{-15}} \approx 4.0\times 10^{14}. $$(A1)
因斜率等于 $GM$：$M = \dfrac{\text{斜率}}{G}$。(M1)
$$ M = \frac{4.0\times 10^{14}}{6.67\times 10^{-11}} \approx 6.0\times 10^{24}\ \mathrm{kg}. $$(A1)
场强度是势的负梯度，$g = -\dfrac{\Delta V_g}{\Delta r}$。(A1)
故在 $V_g$ 对 $r$ 的图上，某点的场强度等于曲线斜率（局部梯度）的相反数。(A1)
负号使 $g$ 指向势减小方向，即指向行星内部。(R1)
Geostationary satellite, $T = 8.64\times 10^{4}\ \mathrm{s}$, Earth $M = 6.0\times 10^{24}\ \mathrm{kg}$, $R = 6.4\times 10^{6}\ \mathrm{m}$. (a) two conditions; (b) show $r = (GMT^{2}/4\pi^{2})^{1/3}$ and find it; (c) altitude; (d) orbital speed; (e) field strength at the orbit vs surface.地球静止卫星，$T = 8.64\times 10^{4}\ \mathrm{s}$，地球 $M = 6.0\times 10^{24}\ \mathrm{kg}$，$R = 6.4\times 10^{6}\ \mathrm{m}$。(a) 两个条件；(b) 证明 $r = (GMT^{2}/4\pi^{2})^{1/3}$ 并求值；(c) 高度；(d) 轨道速度；(e) 轨道处场强与表面比较。
The period must equal one sidereal day (matching Earth's rotation), so $T = 8.64\times 10^{4}\ \mathrm{s}$, and the satellite must orbit eastward in the equatorial plane. (B1)
The orbit must be circular about Earth's centre so that the satellite stays fixed above one point on the equator. (B1)
Gravity supplies the centripetal force, and with $v = \dfrac{2\pi r}{T}$ this gives Kepler's third law $T^{2} = \dfrac{4\pi^{2}}{GM}r^{3}$. (M1)
Rearranging for $r$: $r = \left(\dfrac{GMT^{2}}{4\pi^{2}}\right)^{1/3}$. (A1)
$$ r = \left(\frac{(6.67\times 10^{-11})(6.0\times 10^{24})(8.64\times 10^{4})^{2}}{4\pi^{2}}\right)^{1/3}. $$(M1 for substitution)
$$ r = \left(7.57\times 10^{22}\right)^{1/3} \approx 4.2\times 10^{7}\ \mathrm{m}. $$(A1)
Altitude is the orbital radius minus Earth's radius: $h = r - R = 4.23\times 10^{7} - 6.4\times 10^{6}$. (M1)
$$ h \approx 3.6\times 10^{7}\ \mathrm{m}. $$(A1)
Use $v = \dfrac{2\pi r}{T} = \dfrac{2\pi (4.23\times 10^{7})}{8.64\times 10^{4}}$. (M1)
$$ v \approx 3.1\times 10^{3}\ \mathrm{m\,s^{-1}}. $$(A1)
$g = \dfrac{GM}{r^{2}} = \dfrac{(6.67\times 10^{-11})(6.0\times 10^{24})}{(4.23\times 10^{7})^{2}} \approx 0.22\ \mathrm{N\,kg^{-1}}$. (M1)
The surface value is $g_{0} = GM/R^{2} \approx 9.8\ \mathrm{N\,kg^{-1}}$, so the orbital field is about $0.22/9.8 \approx 1/44$ of the surface value, consistent with the inverse-square fall-off over a $\approx 6.6$-fold increase in $r$. (A1)
周期须等于一个恒星日（与地球自转一致），即 $T = 8.64\times 10^{4}\ \mathrm{s}$，且卫星须在赤道平面内自西向东运行。(B1)
轨道须为绕地心的圆，使卫星固定在赤道上某点正上方。(B1)
引力提供向心力，结合 $v = \dfrac{2\pi r}{T}$ 得开普勒第三定律 $T^{2} = \dfrac{4\pi^{2}}{GM}r^{3}$。(M1)
对 $r$ 整理：$r = \left(\dfrac{GMT^{2}}{4\pi^{2}}\right)^{1/3}$。(A1)
$$ r = \left(\frac{(6.67\times 10^{-11})(6.0\times 10^{24})(8.64\times 10^{4})^{2}}{4\pi^{2}}\right)^{1/3}. $$(代入得 M1)
$$ r = \left(7.57\times 10^{22}\right)^{1/3} \approx 4.2\times 10^{7}\ \mathrm{m}. $$(A1)
高度为轨道半径减地球半径：$h = r - R = 4.23\times 10^{7} - 6.4\times 10^{6}$。(M1)
$$ h \approx 3.6\times 10^{7}\ \mathrm{m}. $$(A1)
用 $v = \dfrac{2\pi r}{T} = \dfrac{2\pi (4.23\times 10^{7})}{8.64\times 10^{4}}$。(M1)
$$ v \approx 3.1\times 10^{3}\ \mathrm{m\,s^{-1}}. $$(A1)
$g = \dfrac{GM}{r^{2}} = \dfrac{(6.67\times 10^{-11})(6.0\times 10^{24})}{(4.23\times 10^{7})^{2}} \approx 0.22\ \mathrm{N\,kg^{-1}}$。(M1)
表面值为 $g_{0} = GM/R^{2} \approx 9.8\ \mathrm{N\,kg^{-1}}$，故轨道处场约为表面值的 $0.22/9.8 \approx 1/44$，与 $r$ 增大约 $6.6$ 倍下的反平方衰减一致。(A1)
Satellite $m = 2500\ \mathrm{kg}$, $r_{1} = 7.0\times 10^{6}\ \mathrm{m}$, raised to $r_{2} = 1.4\times 10^{7}\ \mathrm{m}$, Earth $M = 6.0\times 10^{24}\ \mathrm{kg}$. (a) show $E = -GMm/2r$; (b) energy in the lower orbit; (c) energy to raise it; (d) why a higher orbit is slower despite added energy.卫星 $m = 2500\ \mathrm{kg}$，$r_{1} = 7.0\times 10^{6}\ \mathrm{m}$，抬升至 $r_{2} = 1.4\times 10^{7}\ \mathrm{m}$，地球 $M = 6.0\times 10^{24}\ \mathrm{kg}$。(a) 证明 $E = -GMm/2r$；(b) 较低轨道能量；(c) 抬升所需能量；(d) 为何补充能量后更高轨道反而更慢。
For a circular orbit gravity is the centripetal force, $\dfrac{GMm}{r^{2}} = \dfrac{mv^{2}}{r}$, so $mv^{2} = \dfrac{GMm}{r}$ and the kinetic energy is $E_k = \tfrac{1}{2}mv^{2} = \dfrac{GMm}{2r}$. (M1)
The potential energy is $E_p = -\dfrac{GMm}{r}$. (M1)
$$ E = E_k + E_p = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}. $$(A1)
$E_{1} = -\dfrac{GMm}{2r_{1}} = -\dfrac{(6.67\times 10^{-11})(6.0\times 10^{24})(2500)}{2(7.0\times 10^{6})}$. (M1)
$$ E_{1} \approx -7.1\times 10^{10}\ \mathrm{J}. $$(A1)
The energy in the higher orbit is $E_{2} = -\dfrac{GMm}{2r_{2}} \approx -3.6\times 10^{10}\ \mathrm{J}$. (M1)
The energy that must be supplied is the difference: (M1)
$$ \Delta E = E_{2} - E_{1} = (-3.57\times 10^{10}) - (-7.14\times 10^{10}) \approx +3.6\times 10^{10}\ \mathrm{J}. $$(A1)
Adding energy makes the total $E = -GMm/2r$ less negative, which requires a larger $r$. (A1)
But the kinetic energy $E_k = GMm/2r$ also falls as $r$ grows, so the orbital speed $v = \sqrt{GM/r}$ decreases. The supplied energy goes into potential energy (which rises by $2|\Delta E_k|$), more than offsetting the drop in kinetic energy. (R1)
圆轨道中引力即向心力，$\dfrac{GMm}{r^{2}} = \dfrac{mv^{2}}{r}$，故 $mv^{2} = \dfrac{GMm}{r}$，动能 $E_k = \tfrac{1}{2}mv^{2} = \dfrac{GMm}{2r}$。(M1)
势能为 $E_p = -\dfrac{GMm}{r}$。(M1)
$$ E = E_k + E_p = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}. $$(A1)
$E_{1} = -\dfrac{GMm}{2r_{1}} = -\dfrac{(6.67\times 10^{-11})(6.0\times 10^{24})(2500)}{2(7.0\times 10^{6})}$。(M1)
$$ E_{1} \approx -7.1\times 10^{10}\ \mathrm{J}. $$(A1)
较高轨道能量为 $E_{2} = -\dfrac{GMm}{2r_{2}} \approx -3.6\times 10^{10}\ \mathrm{J}$。(M1)
须提供的能量为二者之差：(M1)
$$ \Delta E = E_{2} - E_{1} = (-3.57\times 10^{10}) - (-7.14\times 10^{10}) \approx +3.6\times 10^{10}\ \mathrm{J}. $$(A1)
补充能量使总能量 $E = -GMm/2r$ 变得不那么负，这要求更大的 $r$。(A1)
但动能 $E_k = GMm/2r$ 也随 $r$ 增大而减小，故轨道速度 $v = \sqrt{GM/r}$ 下降。补充的能量进入势能（其上升量为 $2|\Delta E_k|$），足以超过动能的下降。(R1)
Probe $1200\ \mathrm{kg}$ launched at $1.0\times 10^{4}\ \mathrm{m\,s^{-1}}$ from a planet $M = 4.9\times 10^{24}\ \mathrm{kg}$, $R = 6.1\times 10^{6}\ \mathrm{m}$. (a) escape speed; (b) total energy at launch, bound or unbound; (c) if bound, the maximum distance from the centre.探测器 $1200\ \mathrm{kg}$ 以 $1.0\times 10^{4}\ \mathrm{m\,s^{-1}}$ 从行星 $M = 4.9\times 10^{24}\ \mathrm{kg}$，$R = 6.1\times 10^{6}\ \mathrm{m}$ 发射。(a) 逃逸速度；(b) 发射时总能量，束缚还是非束缚；(c) 若束缚，到中心的最大距离。
$v_{esc} = \sqrt{\dfrac{2GM}{R}} = \sqrt{\dfrac{2(6.67\times 10^{-11})(4.9\times 10^{24})}{6.1\times 10^{6}}}$. (M1)
$$ v_{esc} = \sqrt{1.07\times 10^{8}} \approx 1.0\times 10^{4}\ \mathrm{m\,s^{-1}}. $$(A1)
Total mechanical energy at launch is $E = \tfrac{1}{2}mv^{2} - \dfrac{GMm}{R}$. (M1)
$$ E = \tfrac{1}{2}(1200)(1.0\times 10^{4})^{2} - \frac{(6.67\times 10^{-11})(4.9\times 10^{24})(1200)}{6.1\times 10^{6}}. $$ $$ E = 6.00\times 10^{10} - 6.43\times 10^{10} \approx -4.3\times 10^{9}\ \mathrm{J}. $$(A1)
Since $E < 0$ the probe is bound: it rises, slows to rest at a maximum distance, and falls back. The launch speed $1.0\times 10^{4}$ is just below the escape speed, so it does not escape. (R1)
At the highest point $v = 0$, so all the energy is potential: $E = -\dfrac{GMm}{r_{\max}}$. (M1)
$$ r_{\max} = -\frac{GMm}{E} = -\frac{(6.67\times 10^{-11})(4.9\times 10^{24})(1200)}{-4.29\times 10^{9}}. $$(M1)
$$ r_{\max} \approx 9.1\times 10^{7}\ \mathrm{m}. $$(A1)
$v_{esc} = \sqrt{\dfrac{2GM}{R}} = \sqrt{\dfrac{2(6.67\times 10^{-11})(4.9\times 10^{24})}{6.1\times 10^{6}}}$。(M1)
$$ v_{esc} = \sqrt{1.07\times 10^{8}} \approx 1.0\times 10^{4}\ \mathrm{m\,s^{-1}}. $$(A1)
发射时总机械能为 $E = \tfrac{1}{2}mv^{2} - \dfrac{GMm}{R}$。(M1)
$$ E = \tfrac{1}{2}(1200)(1.0\times 10^{4})^{2} - \frac{(6.67\times 10^{-11})(4.9\times 10^{24})(1200)}{6.1\times 10^{6}}. $$ $$ E = 6.00\times 10^{10} - 6.43\times 10^{10} \approx -4.3\times 10^{9}\ \mathrm{J}. $$(A1)
因 $E < 0$，探测器为束缚态：它上升、在最大距离处减速至静止，然后回落。发射速率 $1.0\times 10^{4}$ 略低于逃逸速度，故不能逃逸。(R1)
最高点 $v = 0$，故全部能量为势能：$E = -\dfrac{GMm}{r_{\max}}$。(M1)
$$ r_{\max} = -\frac{GMm}{E} = -\frac{(6.67\times 10^{-11})(4.9\times 10^{24})(1200)}{-4.29\times 10^{9}}. $$(M1)
$$ r_{\max} \approx 9.1\times 10^{7}\ \mathrm{m}. $$(A1)