Demonstrating Kepler's Second Law - Equal Areas Swept Out in Equal Times

[UrbanXrisis]

question is http://home.earthlink.net/~urban-xrisis/clip_image002.jpg

I know that an imaginary line adjoining a planet and a sun sweeps out an equal area of space in equal amounts of time.

that means... I know that

$$\frac{1}{2}b(t_2-t_1)=\frac{1}{2}b(t_4-3_1)$$

but I don't know that the question means when it asks to demonstrate that the shaded areas are the same?

 

[SpaceTiger]

that means... I know that

$$\frac{1}{2}b(t_2-t_1)=\frac{1}{2}b(t_4-3_1)$$

but I don't know that the question means when it asks to demonstrate that the shaded areas are the same?

Make sure that when you're calculating the area that it results in units of distance squared (what is the relationship between the time interval, the distance travelled, and the velocity?). Your equation above has units of distance times time.

 

[Hippo]

Just formalize your response.

$$(A_{[1,2]} = A_{[3,4]})$$ $$\leftrightarrow$$ $$(A_{[1,2]} - A_{[3,4]} = 0)$$

So is the latter statement true? Prove it.

And like SpaceTiger wrote, when calculating those areas, don't forget, well, your velocity.

 

[UrbanXrisis]

hmm... all I got is...

$$d=v\Delta t$$
$$\sqrt{b^2+(t_2-t_1)^2}=V_o (t_2-t_1)$$

not sure what to do next...

 

[UrbanXrisis]

well,I see what you are saying now.

$$d=V_o (t_2-t_1)$$
$$A=\frac{1}{2} b V_o(t_2-t_1)$$

this is in terms of m^2

but what does the book mean when it says that I have to deomstrate that the shaded triangles have the same area?

 

[xanthym]

well,I see what you are saying now.

$$d=V_o (t_2-t_1)$$
$$A=\frac{1}{2} b V_o(t_2-t_1)$$

this is in terms of m^2

but what does the book mean when it says that I have to deomstrate that the shaded triangles have the same area?

The book is requesting you prove GEOMETRICALLY that the 2 triangles have equal areas.
Hint: (Triangle Area) = (1/2)*(Altitude)*(Base) for each triangle. Formalize what you've already done.


~~

 

[SpaceTiger]

It looks like you're most of the way there. Incidentally, the reason Kepler's second law applies to this situation is that it represents the limit of a classical unbound orbit as the moving particle's energy goes to infinity (for an arbitrary angular momentum). In other words, the moving particle is going so fast that there is no noticable gravitational deviation about point O.