Motion of Orbiting Satellites · 轨道卫星的运动
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| orbital period/ˈɔːbɪtl ˈpɪərɪəd/ | 轨道周期 | guǐ dào zhōu qī |
The endless fall
- The Moon is always falling toward Earth, yet it never lands.
- A space station circles the planet once every ninety minutes.
- Gravity pulls it inward, but its sideways speed keeps missing the ground.
- One pull, one speed, and a perfect endless curve.
永不停止的下落
- 月球始终在朝地球下落,却从不着陆。
- 空间站每九十分钟绕这颗行星转一圈。
- 引力把它向内拉,但它侧向的速度总是让它"错过"地面。
- 一份拉力、一个速度,合成一条完美的、无尽的曲线。
Gravity is the centripetal pull
- For a circular orbit, gravity supplies exactly the centripetal force:
- Solving for the speed: $v = \sqrt{\dfrac{GM}{r}}$.
- Lower orbits are faster; higher ones are slower.
引力就是向心拉力
- 对于圆轨道,引力恰好提供向心力:
- 解出速度:$v = \sqrt{\dfrac{GM}{r}}$。
- 更低的轨道更快;更高的更慢。
A satellite moves to a lower circular orbit. Its orbital speed... · 一颗卫星变轨至更低的圆形轨道。其轨道速度...
$v = \sqrt{GM/r}$, so a smaller $r$ gives a larger $v$. · $v = \sqrt{GM/r}$，因此更小的$r$导致更大的$v$。
What provides the centripetal force that holds a satellite in a circular orbit? · 是什么提供了将卫星保持在圆形轨道上的向心力？
$GMm/r^2 = mv^2/r$ -- gravity is the centripetal force. · $GMm/r^2 = mv^2/r$ -- 重力是向心力。
Compute the orbital speed where $GM/r = 4.0\times10^7\ \text{m}^2/\text{s}^2$ (in m/s). · 计算轨道速度，其中$GM/r = 4.0\times10^7\ \text{m}^2/\text{s}^2$（单位：m/s）。
$v = \sqrt{4.0\times10^7} \approx 6325\ \text{m/s}$.
Period grows with radius
- Cancelling and rearranging gives the orbital period 轨道周期:
- The period squared grows with the radius cubed -- Kepler's third law.
- A bigger orbit takes much longer to go around.
周期随半径增大
- 消去并整理,得到轨道周期:
- 周期的平方随半径的立方增大——开普勒第三定律。
- 更大的轨道,绕一圈要长得多的时间。
A satellite in a larger orbit has a longer period. · 在更大轨道上的卫星具有更长的周期。
$T^2 \propto r^3$, so a bigger $r$ means a bigger $T$. · $T^2 \propto r^3$，因此更大的$r$意味着更大的$T$。
Faster when closer
- In an elliptical orbit, gravity points along $r$, so it makes no torque.
- With no torque, the planet's angular momentum is conserved.
- It speeds up near the star and sweeps equal areas in equal times.
越近越快
- 在椭圆轨道上,引力沿 $r$ 方向,所以不产生力矩。
- 没有力矩,行星的角动量守恒。
- 它在靠近恒星时加速,并在相等时间内扫过相等面积。
Low orbit or high orbit? · 低轨道还是高轨道？
Gravity provides the centripetal force for a satellite. Sort each fact by orbit height. · 重力为卫星提供向心力。按轨道高度对每个事实进行排序。
In an elliptical orbit, a planet moves fastest... · 在椭圆轨道中，行星运动最快时...
Angular momentum is conserved, so a smaller $r$ means a larger speed -- fastest when closest. · 角动量守恒，因此更小的$r$意味着更快的速度——最近时最快。
A satellite orbits where $GM/r = 5.0\times10^7\ \text{m}^2/\text{s}^2$.
- Orbital speed: $v = \sqrt{GM/r} = \sqrt{5.0\times10^7} \approx 7100\ \text{m/s}$.
- That is about $7.1\ \text{km/s}$ -- typical for a low orbit.
一颗卫星在 $GM/r = 5.0\times10^7\ \text{m}^2/\text{s}^2$ 处运行。
- 轨道速度:$v = \sqrt{GM/r} = \sqrt{5.0\times10^7} \approx 7100\ \text{m/s}$。
- 大约是 $7.1\ \text{km/s}$——低轨道的典型值。
Doubling the satellite's own mass changes its orbital speed at the same radius. · 改变卫星自身的质量会改变其在相同半径处的轨道速度。
The mass $m$ cancels in $GMm/r^2 = mv^2/r$, so $v$ is independent of $m$. · 质量$m$在$GMm/r^2 = mv^2/r$中抵消，因此$v$与$m$无关。
Measure $r$ from the centre of the planet, not the altitude above the surface -- add the planet's radius. The speed formula assumes a circular orbit; an ellipse has a changing speed (fastest at the closest point). And the satellite's own mass $m$ cancels out, so orbital speed and period do not depend on it.
$r$ 要从行星中心量起,而不是地表以上的高度——要加上行星半径。速度公式假设圆轨道;椭圆的速度是变化的(在最近点最快)。而且卫星自身质量 $m$ 会被约掉,所以轨道速度和周期都与它无关。
Gravity is the centripetal force of a circular orbit, giving $v = \sqrt{GM/r}$ (lower means faster) and the orbital period $T^2 = \tfrac{4\pi^2}{GM}r^3$ (Kepler's third law). Because gravity makes no torque, a planet's angular momentum is conserved, so it moves fastest at its closest approach. Measure $r$ from the centre; the orbiting mass cancels.
引力是圆轨道的向心力,给出 $v = \sqrt{GM/r}$(越低越快)和轨道周期 $T^2 = \tfrac{4\pi^2}{GM}r^3$(开普勒第三定律)。因为引力不产生力矩,行星的角动量守恒,所以它在最近点运动最快。$r$ 从中心量起;绕行的质量会被约掉。