In basic high school physics/calculus you learn that you can formulate equations for velocity and displacement under constant acceleration as: a ( t ) = a 0 v ( t ) = a 0 t + v 0 x ( t ) = 1 2 a 0 t 2 + v 0 t + x 0 My question is how would you formulate a similar equation when acceleration is dependent on the inverse-square of distance from a point, such as Coulomb's Law or Isaac Newton's inverse-square law of universal gravitation?

Question

In basic high school physics/calculus you learn that you can formulate equations for velocity and displacement under constant acceleration as:  a  (  t  )  =      a    0    v  (  t  )  =      a    0    t  +      v    0    x  (  t  )  =      1    2        a    0        t    2    +      v    0    t  +      x    0  My question is how would you formulate a similar equation when acceleration is dependent on the inverse-square of distance from a point, such as Coulomb&#039;s Law or Isaac Newton&#039;s inverse-square law of universal gravitation?

nutnhonyl8 · Accepted Answer

Since   a  =            d      v              d      t        =            d      x              d      t                  d      v              d      x        =  v            d      v              d      x        =      d          d      x        (      1    2        v    2    ), integration gives       v    2    =      v    ∞    2    −            2      k        x  . Then   t  =  ∫            d      x        v    =  ∫            d      x                      v        ∞        2            −                        2          k                x            . If you evaluate that integral (hint: substitute   x  =            2      k              v      ∞      2            csc    2    ⁡  θ), you&#039;ll have t as a function of x, not the other way round.

What are the scalar equations displacement of acceleration obeys the inverse-square law?

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