- #1
IAmBadAtMath
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- Homework Statement
- A meteroide travelling at earth at a speed of 8.5km/s. It is currently 12R (earth radius) from the center of the Earth. At which speed does the meteroite crash into Earth?
- Relevant Equations
- Earths Radius (R) = 6,4*10^24m
Mass of earth (m2): 6*10^24kg
Mass of meteroide (m1): ?
𝜸 = Gravitational constant (6.67408*10^-11Nm^2/kg^2)
r: radius( in this case 12R)
Fg = 𝜸(m1*m2)/r^2
F=m*a
I haven't gotten anywhere. I don't find it possible to calculate this since Fg varies based on the Mass of the meteroide and because of that it will change acceleration. I thought about trying to remove m1 by making F=m*a the same as 𝜸(m1*m2)/r^2 since I think they are the same force.
m*a= 𝜸(m1*m2)/r^2
a = 𝜸*m2/r^2
I then thought about getting the avarage acceleration out of it since its acceleration will increase the closer it gets to earth
𝜸*6*10^24kg/(12* 6.4*10^24m)^2 = 0.068m/s^2
gravity at the surface of the Earth is 9.81 m/s^2
(0.068m/s^2 + 9.81m/s^2)/2 = 4.9m/s^2
At this point i don't know what I can do anymore. I have what I think is the avarage acceleration for the meteroide, but I don't know what to do next and I honestly don't even know If I have calculated right at all.
m*a= 𝜸(m1*m2)/r^2
a = 𝜸*m2/r^2
I then thought about getting the avarage acceleration out of it since its acceleration will increase the closer it gets to earth
𝜸*6*10^24kg/(12* 6.4*10^24m)^2 = 0.068m/s^2
gravity at the surface of the Earth is 9.81 m/s^2
(0.068m/s^2 + 9.81m/s^2)/2 = 4.9m/s^2
At this point i don't know what I can do anymore. I have what I think is the avarage acceleration for the meteroide, but I don't know what to do next and I honestly don't even know If I have calculated right at all.