Calculate gravitational acceleration without mass of both objects

Click For Summary
Calculating gravitational acceleration without knowing the mass of both objects presents challenges, as gravitational force (Fg) depends on mass. The user attempts to relate force and acceleration through the equations F = m*a and Fg = 𝜸(m1*m2)/r^2, leading to an expression for acceleration. They calculate an average acceleration of 4.9 m/s^2 for a meteoroid approaching Earth, but express uncertainty about the accuracy of their calculations. The discussion hints at the conservation of energy as a potential method to relate speeds at different positions. The conversation emphasizes the complexities of gravitational interactions and the need for further clarification on the calculations.
IAmBadAtMath
Messages
2
Reaction score
0
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.
 
Physics news on Phys.org
You are trying to relate a known speed at position 12 R with an unknown speed at position R. What kind of equation would do that? Hint: What quantity is conserved?
 
Thread 'Chain falling out of a horizontal tube onto a table'
My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

Similar threads

Replies
16
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 33 ·
2
Replies
33
Views
3K
  • · Replies 20 ·
Replies
20
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
25
Views
3K
Replies
17
Views
2K
  • · Replies 3 ·
Replies
3
Views
831