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Gravitation — Physics STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 SciencePhysicsGravitation5 Marks
Question
A solid sphere of mass m and radius r is placed inside a hollow thin spherical shell of mass M and radius R as shown in A particle of mass m' is placed on the line joining the two centres at a distance x from the point of contact of the sphere and the shell. Find the magnitude of the resultant gravitational force on this particle due to the sphere and the shell if,
  1. r < x < 2r.
  2. 2r < x < 2R.
  3. x > 2R.

✓

Answer

  1. m' is placed at a distance x from ‘O’.

If r < x, 2r, Let’s consider a thin shell of man

$\text{dm}=\frac{\text{m}}{\big(\frac{4}{3}\big)\pi\text{r}^2}\times\frac{4}{3}\pi\text{x}^3=\frac{\text{mx}^3}{\text{r}^3}$

Thus $\int\text{dm}=\frac{\text{mx}^3}{\text{r}^3}$

Then gravitational force $\text{F}=\frac{\text{Gmdm}}{\text{x}^2}=\frac{\frac{\text{Gmx}^3}{\text{r}^3}}{\text{x}^2}=\frac{\text{Gmx}}{\text{r}^3}$

  1. x < 2R, then F is due to only the sphere.

$\text{F}=\frac{\text{Gmm'}}{(\text{x}-\text{r})^2}$

  1. If x > 2R, then Gravitational force is due to both sphere & shell, then due to shell,

$\text{F}=\frac{\text{GMm'}}{(\text{x}-\text{R})^2}$

due to the sphere $=\frac{\text{Gmm'}}{(\text{x}-\text{r})^2}$

So, Resultant force $=\frac{\text{Gmm'}}{(\text{x}-\text{r})^2}+\frac{\text{GMm'}}{(\text{x}-\text{R})^2}$

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