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If you leave a ball weighing 10 kg at a height of 500 km above sea level (neglecting air friction). How can calculate how long the ball hits the ground and what will be its speed? enter image description here

I know that: On earth I can calculate the time taken to the object to fall by this equivalent:

x = x0 + v0t + (at^2)/2
x = 0
x0 = 500000
v0 = 0
a = -g = 10 m/s^2
0 = 500000 - 5t^2
t = 316.227sec

Also I know how to calculate the force between objects:

g = GMm/r^2
jinawee
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Zilberman Rafael
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  • Well, $F=-\frac{GMm}{r^2}$ and $F=ma$ so that is a starting point. Have you tried anything yet? – jerk_dadt Mar 24 '14 at 18:02
  • @jerk_dadt This is not what I'm learning in school so a detailed explanation would really help, still have not tried anything. – Zilberman Rafael Mar 24 '14 at 18:04
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    See this Wikipedia article – John Rennie Mar 24 '14 at 18:05
  • @JohnRennie I did not understand a single word. – Zilberman Rafael Mar 24 '14 at 18:09
  • Can you edit your question to give us some idea what you already know about calculating equations of motion. I'm afraid that if the answer is "not much" you have a long haul ahead of you. The Wikipedia article gives an equation for the time as a function of distance fallen. Take that equation, put $y_0$ as the initial height (NB distance from the centre of the Earth) and $y$ as the final height (again from the centre so the final height is the radius of the Earth) and the equation will give you the time taken. – John Rennie Mar 24 '14 at 18:19
  • There is a related question here and the derivation of the equation is given in an answer to this question. However you may find this heavy going. – John Rennie Mar 24 '14 at 18:20
  • The working you've shown is correct, but it assumes the gravitational acceleration is constant. However 500km is far enough for the gravitational acceleration to change quite a bit so the fall time is longer than your calculation. The complicated equation in the Wiki article takes this into account. BTW using the full equaiton I get 340 seconds for the fall time. – John Rennie Mar 24 '14 at 18:28
  • @JohnRennie I hate coping from other people. I want to know how to solve this. – Zilberman Rafael Mar 24 '14 at 18:29
  • Hi Zilberman Rafael. If you haven't already done so, please take a minute to read the definition of when to use the homework tag, and the Phys.SE policy for homework-like problems. – Qmechanic Mar 24 '14 at 18:31
  • @ZilbermanRafael: the derivation is given in an answer to this question. It's hard going though. – John Rennie Mar 24 '14 at 18:32
  • @JohnRennie There is no way you can at least explain to me what's going on there? – Zilberman Rafael Mar 24 '14 at 18:37
  • I suspect that the level of the course that the OP is taking is non-calculus. Another hint of that is $g=10$. It's possible that in the original question it is assumed that gravity is constant. Perhaps @ZilbermanRafael can tell us if the course has dealt with non-constant forces, and if it is a calculus-based course. – garyp Mar 24 '14 at 18:45
  • The trouble is the explanation needs to describe how you set up a differential equation to describe the motion of the particle, and how you solve the equation. This isn't as hard as it sounds - in the UK you do it in the last two years at school (age 17-18) and in more detail in the first year of a univversity physics degree. However to understand it requires a fair bit of maths background. To try and explain it here would be a long answer, and probably an unsuccessful one. – John Rennie Mar 24 '14 at 18:47
  • ZilbermanRafael What is happening here is that the acceleration $a$ is changing with $r$ when you go to a large height of 500km. Near the surface, this acceleration is equal to $g=-10m/s^2$ and it is approximately constant for small heights. To solve your problem what you have to do is use $GM/r^2$ as you acceleration, and then proceed with solving the differential equation in $r$ and $t$. As @garyp suggests, this requires a certain level of knowledge in calculus. – udiboy1209 Mar 24 '14 at 18:49
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    possible duplicate of Time taken for object in space to fall to earth – John Alexiou Mar 24 '14 at 18:49
  • @udiboy I'm at 8 grade so I don't know how to solve differential equation (I should learn it in the next month). – Zilberman Rafael Mar 24 '14 at 18:56
  • 8th grade is around age 14 isn't it? I couldn't have solved the differential equation describing this motion until around age 18. Good luck, but I suspect your ambition is running away with you. – John Rennie Mar 24 '14 at 19:11
  • Also, his analysis is spot on given the simplifying approximation of constant acceleration (even though the approximation is really not a good one). @JohnRennie in the US only the top 5% or so of 18 year olds would be able to solve this. – garyp Mar 24 '14 at 19:17

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