An object is dropped from distance d=1m about the planet A's surface, how much time does it take to land?

Question

An object is dropped from distance d=1m about the planet A's surface, how much time does it take to land?
take the mass of the planet to be 1.99x10^30kg. Radius of planet= 6.37x10^6m.

The answer given by our teacher is 7.8x10^-4s but my answer is about half of that.
I used P.E. at 1m= P.E. at surface of planet+ K.E. at surface of planet
then used time=distance/ velocity
so what is wrong?

m=meters
the whole dropping happened on planet A, an imaginary planet.

thanks, i got the answer already!

Answer 1

Your teacher is right
Accleration due to gravity = GM/r^2 = g
G=7.7481*10^-5
M=1.99x10^30kg
r= 6.37x10^6m

g=3.72*10^6
initial velocity = 0m/s
s=0.5*g*t^2
1=0.5*3.72*10^6*t^2
t=sqrt(6.111*10^-7) = 7.817*10^-4

By your method:
P.E. at 1m= mg*1=mg
P.E. at surface of planet= 0
K.E. at surface of planet = 0.5mv^2
mg = 0.5mv^2
v^2 = sqrt(2*g)
anyway you need to calculate g at planet A
Besides its a constant accleration problem , YOU CAN'T USE time=distance/ velocity
USE THE SECOND EQUATION OF KINEMATICS:

s = u*t + 0.5*a*t^2
s=distance
u=initial velocity
t=time
a=acceleration

Answer 2

relies upon on key complication-unfastened enter components which includes: does m=miles or meters , and which planet? (gravity differs). And interpreting your preliminary question, what does the solar could desire to do with something? The moon might instruct to be greater of an interacting factor if something.... yet....regardless of...

Answer 3

Try using
time = distance / (AVERAGE velocity)

The velocity starts out at zero, and ends up at the velocity you calculated from energy conservation. The average velocity is one-half of what you calculated.

Hope this helps.