91影视

1. Radius of the lead sphere is$R=4.00\text{cm}$
2. Mass of the lead sphere before hollowing is$M=2.95\text{kg}$
3. Mass of the small lead sphere is$m=0.431\text{kg}$
4. Distance of the small sphere from the centre of lead sphere is$d=9.00\text{cm}$
5. Gravitational constant,$G=6.67脳{10}^{鈭�11}\text{鈥凬}鈰�{\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$

Newton鈥檚 law of gravitation states that any particle in the universe attracts any other particle with a gravitational force whose magnitude is

$F=G\frac{m_1{m}_2}{r^2}$

Here, $m_1$and $m_2$are masses of the particles and $r$is their separation and$G$is the gravitational constant.

We can use the concept of the force of gravitation and density to find the force between the hollowed sphere and the small sphere. For this, we need to find the mass of the sphere which can fill the cavity. Then, we subtract the force between the mass of the cavity and the small mass from the force between the whole sphere and the small sphere. This would give us the required answer.

Formulae:

a) Density of a material, $蚁=\frac{M}{V}$where M is mass of the material and V is the volume of the material.

b) Gravitational force, $F=\frac{GMm}{d^2}$ ...(i)

If the lead sphere was not hollowed, then the force on the small sphere would be $F_1=\frac{GMm}{d^2}$ ...(ii)

But if the sphere is hollowed, then cavity has radius$r=\frac{R}{2}$then material that fills the cavity will have same density to that of lead sphere of radius R.

$\begin{array}{c}{蚁}_c=蚁\\ \frac{M_c}{\frac{4}{3}蟺r^3}=\frac{M}{\frac{4}{3}蟺R^3}\\ \end{array}$

We can cancel the same terms from the denominator, hence we get

$\frac{M_c}{r^3}=\frac{M}{R^3}$

where,$M_c$is mass of cavity that fills the material.

We can write the above equation for$M_c$,

$M_c=\left(\frac{r^3}{R^3}\right)M$

Substituting the value of r , we get

$\begin{array}{c}{M}_c=\left(\frac{{\left(\frac{R}{2}\right)}^3}{R^3}\right)M\\ =\frac{M}{8}\end{array}$

$M_c=\frac{M}{8}$

The centre of cavity can be found as:,$d鈭�r=d鈭�\left(\frac{R}{2}\right)$which is the distance of mass m from the cavity

So, the force between mass of material filled in cavity and small mass at distance
.$d鈭�\frac{R}{2}$is

$\begin{array}{c}{F}_2=\frac{G\left(M_c\right)m}{{\left(d鈭�\frac{R}{2}\right)}^2}\\ =\frac{G\left(\frac{M}{8}\right)m}{{\left(d鈭�\frac{R}{2}\right)}^2}\end{array}$

The force between hollow sphere and the mass m after substituting the values from equations (ii) & (iii) is

$\begin{array}{c}F=F_1鈭�F_2\\ =\frac{GMm}{d^2}鈭�\frac{G\left(\frac{M}{8}\right)m}{{\left(d鈭�\frac{R}{2}\right)}^2}\\ =GMm\left(\frac{1}{d^2}鈭�\frac{\left(\frac{1}{8}\right)}{{\left(d鈭�\frac{R}{2}\right)}^2}\right)\\ =GMm\left(\frac{1}{d^2}鈭�\frac{1}{8{\left(d鈭�\frac{R}{2}\right)}^2}\right)\\ =\frac{GMm}{d^2}\left(1鈭�\frac{1}{8{\left(1鈭�\frac{R}{2d}\right)}^2}\right)\end{array}$

Substituting all the given values, we get

$\begin{array}{c}F=\frac{6.67脳{10}^{鈭�11}\text{鈥凬}鈰�{\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}脳2.95\text{鈥刱驳}脳0.431\text{鈥刱驳}}{{\left(9脳{10}^{鈭�2}\text{鈥刴}\right)}^2}\left(1鈭�\frac{1}{8{\left[1鈭�\frac{4脳{10}^{鈭�2}\text{鈥刴}}{18脳{10}^{鈭�2}\text{鈥刴}}\right]}^2}\right)\\ =8.306脳{10}^{鈭�9}\text{N}\end{array}$

The net gravitational force is .$8.306脳{10}^{鈭�9}\text{鈥凬}$