Integral of a vector difference
Not sure how to deal with this problem:
$$dfrac {doverrightarrow {h}}{dt}=overrightarrow {r_1}times left( -dfrac {m}{left( overrightarrow {r_2}-overrightarrow {r_1}right) ^{2}}right)$$
where $r_2$ and $r_1$ are vectors. The parameter $t$ represents time. $h$ is the specific angular momentum. The vector direction of the right side term, I suppose, would be $(r_2-r_1)$.
This is as far as I get. But I’m not sure if I’m headed in the right direction:
$$dfrac {doverrightarrow {h}}{dt}=overrightarrow {r_1}times left( -dfrac {m}{left( overrightarrow {r_2}-overrightarrow {r_1}right) ^{3}}left( overrightarrow {r_2}-overrightarrow {r_1}right) right)$$
$$int dfrac {doverrightarrow {h}}{dt} dt=int -dfrac {m}{left( overrightarrow {r_2}-overrightarrow {r_1}right) ^{3}}r_2r_1sin theta dt$$
$$h=int dfrac {mr_2r_1sin theta }{left( left( r_{2}cos theta _{2}-r_{1}cos theta _{1}right) +left( r_{2}sin theta _{2}-r_{1}sin theta _{1}right) right) ^{frac {3}{2}}} dt$$
$$h=dfrac {mr_{2}r_{1}sin theta }{left( overrightarrow {r}_{2}-overrightarrow {r}_{1}right) ^{3}}t+C$$
calculus
asked Nov 29 '18 at 3:17
alexmarisonalexmarison
62
add a comment |
Not sure how to deal with this problem:
$$dfrac {doverrightarrow {h}}{dt}=overrightarrow {r_1}times left( -dfrac {m}{left( overrightarrow {r_2}-overrightarrow {r_1}right) ^{2}}right)$$
where $r_2$ and $r_1$ are vectors. The parameter $t$ represents time. $h$ is the specific angular momentum. The vector direction of the right side term, I suppose, would be $(r_2-r_1)$.
This is as far as I get. But I’m not sure if I’m headed in the right direction:
$$dfrac {doverrightarrow {h}}{dt}=overrightarrow {r_1}times left( -dfrac {m}{left( overrightarrow {r_2}-overrightarrow {r_1}right) ^{3}}left( overrightarrow {r_2}-overrightarrow {r_1}right) right)$$
$$int dfrac {doverrightarrow {h}}{dt} dt=int -dfrac {m}{left( overrightarrow {r_2}-overrightarrow {r_1}right) ^{3}}r_2r_1sin theta dt$$
$$h=int dfrac {mr_2r_1sin theta }{left( left( r_{2}cos theta _{2}-r_{1}cos theta _{1}right) +left( r_{2}sin theta _{2}-r_{1}sin theta _{1}right) right) ^{frac {3}{2}}} dt$$
$$h=dfrac {mr_{2}r_{1}sin theta }{left( overrightarrow {r}_{2}-overrightarrow {r}_{1}right) ^{3}}t+C$$
calculus
asked Nov 29 '18 at 3:17
alexmarisonalexmarison
62
add a comment |
Not sure how to deal with this problem:
$$dfrac {doverrightarrow {h}}{dt}=overrightarrow {r_1}times left( -dfrac {m}{left( overrightarrow {r_2}-overrightarrow {r_1}right) ^{2}}right)$$
where $r_2$ and $r_1$ are vectors. The parameter $t$ represents time. $h$ is the specific angular momentum. The vector direction of the right side term, I suppose, would be $(r_2-r_1)$.
This is as far as I get. But I’m not sure if I’m headed in the right direction:
$$dfrac {doverrightarrow {h}}{dt}=overrightarrow {r_1}times left( -dfrac {m}{left( overrightarrow {r_2}-overrightarrow {r_1}right) ^{3}}left( overrightarrow {r_2}-overrightarrow {r_1}right) right)$$
$$int dfrac {doverrightarrow {h}}{dt} dt=int -dfrac {m}{left( overrightarrow {r_2}-overrightarrow {r_1}right) ^{3}}r_2r_1sin theta dt$$
$$h=int dfrac {mr_2r_1sin theta }{left( left( r_{2}cos theta _{2}-r_{1}cos theta _{1}right) +left( r_{2}sin theta _{2}-r_{1}sin theta _{1}right) right) ^{frac {3}{2}}} dt$$
$$h=dfrac {mr_{2}r_{1}sin theta }{left( overrightarrow {r}_{2}-overrightarrow {r}_{1}right) ^{3}}t+C$$
calculus
asked Nov 29 '18 at 3:17
alexmarisonalexmarison
62
Not sure how to deal with this problem:
$$dfrac {doverrightarrow {h}}{dt}=overrightarrow {r_1}times left( -dfrac {m}{left( overrightarrow {r_2}-overrightarrow {r_1}right) ^{2}}right)$$
where $r_2$ and $r_1$ are vectors. The parameter $t$ represents time. $h$ is the specific angular momentum. The vector direction of the right side term, I suppose, would be $(r_2-r_1)$.
This is as far as I get. But I’m not sure if I’m headed in the right direction:
$$dfrac {doverrightarrow {h}}{dt}=overrightarrow {r_1}times left( -dfrac {m}{left( overrightarrow {r_2}-overrightarrow {r_1}right) ^{3}}left( overrightarrow {r_2}-overrightarrow {r_1}right) right)$$
$$int dfrac {doverrightarrow {h}}{dt} dt=int -dfrac {m}{left( overrightarrow {r_2}-overrightarrow {r_1}right) ^{3}}r_2r_1sin theta dt$$
$$h=int dfrac {mr_2r_1sin theta }{left( left( r_{2}cos theta _{2}-r_{1}cos theta _{1}right) +left( r_{2}sin theta _{2}-r_{1}sin theta _{1}right) right) ^{frac {3}{2}}} dt$$
$$h=dfrac {mr_{2}r_{1}sin theta }{left( overrightarrow {r}_{2}-overrightarrow {r}_{1}right) ^{3}}t+C$$
calculus
calculus
asked Nov 29 '18 at 3:17
alexmarisonalexmarison
62
asked Nov 29 '18 at 3:17
alexmarisonalexmarison
62
edited Nov 29 '18 at 23:38
alexmarison
asked Nov 29 '18 at 3:17
alexmarisonalexmarison
62
asked Nov 29 '18 at 3:17
alexmarisonalexmarison
62
asked Nov 29 '18 at 3:17
alexmarisonalexmarison
62
62
add a comment |
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