Planet X has a gravitational field that decays exponentially with distance from the planet’s
centre, so that the force acting on a particle P of mass m is –mge^(-γr)ȓ, where g > 0 and γ> 0 are
constants and the distance from the centre, r, is measured in the direction of the unit vector ȓ.
Use Newton’s second law F = ma to show that the equation governing the motion of the
particle P near the planet is
r’’ = −ge(−γr).
NOTE: r’’ = r dot dot
do i equate the total force to ma i.e
-mge^(-γr)ȓ=ma
-----------------
My attempted solution:
divide both sides by m
-ge^(-γr)ȓ=a
then i dont know what to do
-mge^(-γr)ȓ=ma
centre, so that the force acting on a particle P of mass m is –mge^(-γr)ȓ, where g > 0 and γ> 0 are
constants and the distance from the centre, r, is measured in the direction of the unit vector ȓ.
Use Newton’s second law F = ma to show that the equation governing the motion of the
particle P near the planet is
r’’ = −ge(−γr).
NOTE: r’’ = r dot dot
do i equate the total force to ma i.e
-mge^(-γr)ȓ=ma
-----------------
My attempted solution:
divide both sides by m
-ge^(-γr)ȓ=a
then i dont know what to do
-mge^(-γr)ȓ=ma
-
Yes, you are basically correct. But, recall that
r'' = 2nd derivative of r with respect to time, which is just the acceleration.
We are given that the force on P = –mge^(-γr)ȓ,
The presence of the unit vector tells us that this is a vector equation. We can rewrite as a scalar just by dropping the unit vector.
Force on P = ma = - mge^(-yr)
dividing both sides my m gives,
a = - ge^(-yr)
but a = r''
so,
r'' = - g e^(-yr)
r'' = 2nd derivative of r with respect to time, which is just the acceleration.
We are given that the force on P = –mge^(-γr)ȓ,
The presence of the unit vector tells us that this is a vector equation. We can rewrite as a scalar just by dropping the unit vector.
Force on P = ma = - mge^(-yr)
dividing both sides my m gives,
a = - ge^(-yr)
but a = r''
so,
r'' = - g e^(-yr)