Gravitational field problem

  • Thread starter Saitama
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  • #1
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Homework Statement


A thin hemispherical shell of mass M and radius R is placed as shown in figure. The magnitude of gravitational field at P due to the hemispherical shell is ##I_0##. The magnitude of gravitational field at Q due to thin hemispherical shell is given by

A)##I_0/2##

B)##I_0##

C)##\frac{2GM}{9R^2}-I_0##

D)##\frac{2GM}{9R^2}+I_0##


Homework Equations





The Attempt at a Solution


I tried the problem using spherical coordinates and ended up with some messy integrals. Since this is an exam problem, I wonder if I really need to solve those integrals as it would take a lot of time. (I solved the integrals using Wolfram Alpha and the result was not nice so I immediately dropped the approach.) I believe there is a shorter way to solve this.

Any help is appreciated. Thanks!
 

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Answers and Replies

  • #2
Dick
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Homework Helper
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Homework Statement


A thin hemispherical shell of mass M and radius R is placed as shown in figure. The magnitude of gravitational field at P due to the hemispherical shell is ##I_0##. The magnitude of gravitational field at Q due to thin hemispherical shell is given by

A)##I_0/2##

B)##I_0##

C)##\frac{2GM}{9R^2}-I_0##

D)##\frac{2GM}{9R^2}+I_0##


Homework Equations





The Attempt at a Solution


I tried the problem using spherical coordinates and ended up with some messy integrals. Since this is an exam problem, I wonder if I really need to solve those integrals as it would take a lot of time. (I solved the integrals using Wolfram Alpha and the result was not nice so I immediately dropped the approach.) I believe there is a shorter way to solve this.

Any help is appreciated. Thanks!

There is a much faster way. You can picture the field due to the hemisphere as the sum of the field from a whole sphere and the field from an imaginary hemisphere of mass -M covering the lower half of the sphere.
 
  • #3
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92
Hi Dick! :)

There is a much faster way. You can picture the field due to the hemisphere as the sum of the field from a whole sphere and the field from an imaginary hemisphere of mass -M covering the lower half of the sphere.

I consider a sphere of radius R and mass 2M.

The field at P is given by:

$$\frac{2GM}{9R^2}+E_{-M}=I_0=E_M$$

where ##E_{-M}## represents field at P due to the imaginary hemisphere of mass -M and ##E_M## represents the field at P due to hemisphere of mass M.

The field at Q is given by:

$$\frac{2GM}{9R^2}+E'_{-M}$$

where ##E'_{-M}## is the field at Q due to imaginary hemisphere of mass -M.

Since ##E'_{-M}=-E_M=-I_0##, the field at Q is given by:

$$\frac{2GM}{9R^2}-I_0$$

Is this correct?
 
  • #4
Dick
Science Advisor
Homework Helper
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Hi Dick! :)



I consider a sphere of radius R and mass 2M.

The field at P is given by:

$$\frac{2GM}{9R^2}+E_{-M}=I_0=E_M$$

where ##E_{-M}## represents field at P due to the imaginary hemisphere of mass -M and ##E_M## represents the field at P due to hemisphere of mass M.

The field at Q is given by:

$$\frac{2GM}{9R^2}+E'_{-M}$$

where ##E'_{-M}## is the field at Q due to imaginary hemisphere of mass -M.

Since ##E'_{-M}=-E_M=-I_0##, the field at Q is given by:

$$\frac{2GM}{9R^2}-I_0$$

Is this correct?

Hi Pranav-Arora! Yes, that's correct. This sort of a method is called 'using superposition'. For sort of obvious reasons.
 
  • #5
3,816
92
Hi Pranav-Arora! Yes, that's correct. This sort of a method is called 'using superposition'. For sort of obvious reasons.

Yes, I have heard of this method, thanks a lot Dick! :smile:
 

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