磁場中的圓形 --- 電子和太空船的軌道(第三章)

第三章

3.2. 轉彎了 : Norem's 概念

在我們的加速度階段(第一階段)我們的激光強度為L' = 43000 TW. 在1.6 年後，船的速度則為每秒1.32^. 10⁸ ms^-1.加速的距離是4.41^. 10¹⁵ m 長。我們在第一小節內提及(加速 ), 加速船並非朝向 Epsilon Eridani星系加速的，是有一定的角度的---angular Fig.4

在第一章裏，我們已經知道我們能通過充電使用星際磁場：我們可想像光束可被分成幾個部分，在被充電之後會不斷推開彼此。所以充電了的太空船就如一有三垂直軸的系統： x, y和z. 在我們的估量中，是可以維持這麼大的電荷的。在 Fig.5 我們給了一個例子如何把這些變為一束光束。 In the following we assume the interstellar field as orientated perpendicular to the direction Sun-Epsilon Eridani to simplify the scenario.

In Sec.3.3 (Braking) we shall see that the decelerating phase (Stage 4) takes as long as the accelerating phase (Stage 1). If we assume constant acceleration and deceleration, the distance d between Sun and Epsilon Eridani is as same length as the distance of the force-free flight (the green line).

In the last section we already determined the initial velocity at Stage 2 to (see above):

After having been accelerated, our ship overcomes the distance of ten light-years in

Here we see: It must be guaranteed that our brave space pioneers can overcome such a long period in isolation - but this is another theme. Nevertheless, let us concentrate on Stage 3.

To describe the turning we remember the definition of the angular velocity _B , see also (2b.01) ...
(3b.01)
... and the relationship we discovered in Chap.2, Sec.2.1 (go to):
(3b.02)
We realize: the angular velocity depends on the electric charge Q and the mass m of the body - as well as on the magnetic induction B₀. First of all we are interested in the Period T_C of a complete circulation. Please keep in mind that our ship carry out only a part of this circulation. If we equalize both formulas (3b.01) (3b.02), we have:
(3b.03)
How strong is the interstellar magnetic field?

According to Forward, it has a strength of B₀ = 3 ^. 10^-5 g = 3 ^. 10^-9 T [Forward (1964)]. Perhaps this is a bit too optimistic; therefore we would like to refer to newer sources [Musser (2000)], in which a field strength of B₀ = 5 ^. 10^-10 T is given. The next interesting question concerning the ship is: how strong could we charge it up?

Here we have the problem, that the properties of the interstellar matter are not well known. Ionized gas particles could - once the ship is charged - be attracted and captured by the ship's stripes. Thus, the generators on board must compensate this neutralizing effects. We assume that the inert atomic nuclei of the (possible) ionized interstellar gas are not as strong attracted by the ship as the lighter electrons. Therefore we want our ship be charged negatively, perhaps in future it is possible to generate a charge of Q = - 10¹⁰ C.
By the way, we have an additional problem (but here we cannot go into the details): While traveling with relativistic speeds, crew and equipment are being exposed to a high cosmic radiation.

In the previous section we took a ship mass of m' = 7.58 ^. 10⁷ Kg. According to formula (3b.02) we are now able to calculate the angular velocity of the turning to

With this result, our ship needs for one circulation a time T_C of:

With the radius R of the circulation and the angular (see Fig.4) we can calculate the duration T of the turning (Stage 3). To find out the radius R we refer to equation (2b.02):

Therefore we have:

According to Fig.4, tan(/2) is given by

This means an angular of = 2.86^o.

With a few trigonometric calculus and the circulation period T_C we can easily derive the duration T of the turning phase. Here we intent you to calculate the relationship between those periods. Our result:
(3b.04)
If we insert the period T_C into this equation, we get a period of about T = 1.52 a.

Finally, we calculate the length of phase 3 (the blue line in Fig.4) to:
s = v^.T = 0.54 c ^. 1.52 a 0.82 ly ,
referring to a length of nearly s = 7.76 ^. 10¹⁵ m.


	第三章 3.2. 轉彎了 : Norem's 概念在我們的加速度階段(第一階段)我們的激光強度為L' = 43000 TW. 在1.6 年後，船的速度則為每秒1.32^. 10⁸ ms^-1.加速的距離是4.41^. 10¹⁵ m 長。我們在第一小節內提及(加速 ), 加速船並非朝向 Epsilon Eridani星系加速的，是有一定的角度的---angular Fig.4 在第一章裏，我們已經知道我們能通過充電使用星際磁場：我們可想像光束可被分成幾個部分，在被充電之後會不斷推開彼此。所以充電了的太空船就如一有三垂直軸的系統： x, y和z. 在我們的估量中，是可以維持這麼大的電荷的。在 Fig.5 我們給了一個例子如何把這些變為一束光束。 In the following we assume the interstellar field as orientated perpendicular to the direction Sun-Epsilon Eridani to simplify the scenario. In Sec.3.3 (Braking) we shall see that the decelerating phase (Stage 4) takes as long as the accelerating phase (Stage 1). If we assume constant acceleration and deceleration, the distance d between Sun and Epsilon Eridani is as same length as the distance of the force-free flight (the green line). In the last section we already determined the initial velocity at Stage 2 to (see above): After having been accelerated, our ship overcomes the distance of ten light-years in Here we see: It must be guaranteed that our brave space pioneers can overcome such a long period in isolation - but this is another theme. Nevertheless, let us concentrate on Stage 3. To describe the turning we remember the definition of the angular velocity _B , see also (2b.01) ... (3b.01) ... and the relationship we discovered in Chap.2, Sec.2.1 (go to): (3b.02) We realize: the angular velocity depends on the electric charge Q and the mass m of the body - as well as on the magnetic induction B₀. First of all we are interested in the Period T_C of a complete circulation. Please keep in mind that our ship carry out only a part of this circulation. If we equalize both formulas (3b.01) (3b.02), we have: (3b.03) How strong is the interstellar magnetic field? According to Forward, it has a strength of B₀ = 3 ^. 10^-5 g = 3 ^. 10^-9 T [Forward (1964)]. Perhaps this is a bit too optimistic; therefore we would like to refer to newer sources [Musser (2000)], in which a field strength of B₀ = 5 ^. 10^-10 T is given. The next interesting question concerning the ship is: how strong could we charge it up? Here we have the problem, that the properties of the interstellar matter are not well known. Ionized gas particles could - once the ship is charged - be attracted and captured by the ship's stripes. Thus, the generators on board must compensate this neutralizing effects. We assume that the inert atomic nuclei of the (possible) ionized interstellar gas are not as strong attracted by the ship as the lighter electrons. Therefore we want our ship be charged negatively, perhaps in future it is possible to generate a charge of Q = - 10¹⁰ C. By the way, we have an additional problem (but here we cannot go into the details): While traveling with relativistic speeds, crew and equipment are being exposed to a high cosmic radiation. In the previous section we took a ship mass of m' = 7.58 ^. 10⁷ Kg. According to formula (3b.02) we are now able to calculate the angular velocity of the turning to With this result, our ship needs for one circulation a time T_C of: With the radius R of the circulation and the angular (see Fig.4) we can calculate the duration T of the turning (Stage 3). To find out the radius R we refer to equation (2b.02): Therefore we have: According to Fig.4, tan(/2) is given by This means an angular of = 2.86^o. With a few trigonometric calculus and the circulation period T_C we can easily derive the duration T of the turning phase. Here we intent you to calculate the relationship between those periods. Our result: (3b.04) If we insert the period T_C into this equation, we get a period of about T = 1.52 a. Finally, we calculate the length of phase 3 (the blue line in Fig.4) to: s = v^.T = 0.54 c ^. 1.52 a 0.82 ly , referring to a length of nearly s = 7.76 ^. 10¹⁵ m. home \| about \| chapter one \| chapter two \| chapter three \| thinkquest Chapter Three

第三章

3.2. 轉彎了 : Norem's 概念

Chapter Three