The gravitational force between a hollow spherical shell (of radius $$R$$ and uniform density) and a point mass is $$F$$. Show the nature of $$F$$ versus $$r$$ graph where $$r$$ is the distance of the point from the centre of the hollow spherical shell of uniform density.
Consider the diagram, density of the shell is constant.
Let it is $$\rho$$.
$$\begin{aligned} \text { Mass of the shell } & =(\text { density }) \times(\text { volume }) \\ & =(\rho) \times \frac{4}{3} \pi R^3=M \end{aligned}$$
As the density of the shell is uniform, it can be treated as a point mass placed at its centre. Therefore, $$F=$$ gravitational force between $$M$$ and $$m=\frac{G M m}{r^2}$$
$$\begin{aligned} F & =0 \text { for } r< R \quad \text { (i.e., force inside the shell is zero) } \\ & =\frac{G M}{r^2} \text { for } r \geq R \end{aligned}$$
The variation of F versus r is shown in the diagram.
Out of aphelion and perihelion, where is the speed of the earth more and why?
Aphelion is the location of the earth where it is at the greatest distance from the sun and perihelion is the location of the earth where it is at the nearest distance from the sun.
The areal velocity $$\left(\frac{1}{2} \mathbf{r} \times \mathbf{v}\right)$$ of the earth around the sun is constant (Kepler's IInd law). Therefore, the speed of the earth is more at the perihelion than at the aphelion.
What is the angle between the equatorial plane and the orbital plane of
(a) polar satellite?
(b) geostationary satellite?
Consider the diagram where plane of geostationary and polar satellite are shown.
Clearly
(a) Angle between the equatorial plane and orbital plane of a polar satellite is $$90^{\circ}$$.
(b) Angle between equatorial plane and orbital plane of a geostationary satellite is $$0^{\circ}$$.
Mean solar day is the time interval between two successive noon when sun passes through zenith point (meridian).
Sidereal day is the time interval between two successive transit of a distant star through the zenith point (meridian).
By drawing appropriate diagram showing the earth's spin and orbital motion, show that mean solar day is 4 min longer than the sidereal day. In other words, distant stars would rise 4 min early every successive day.
Consider the diagram below, the earth moves from the point P to Q in one solar day.
Every day the earth advances in the orbit by approximately $$1^{\circ}$$. Then, it will have to rotate by $$361^{\circ}$$ (which we define as 1 day) to have the sun at zenith point again.
$$\because 361^{\circ}$$ corresponds to 24 h.
$$\therefore 1^{\circ}$$ corresponds to $$\frac{24}{361} \times 1=0.066 \mathrm{~h}=3.99 \mathrm{~min} \approx 4 \mathrm{~min}$$
Hence, distant stars would rise 4 min early every successive day.
Two identical heavy spheres are separated by a distance 10 times their radius. Will an object placed at the mid-point of the line joining their centres be in stable equilibrium or unstable equilibrium? Give reason for your answer.
Let the mass and radius of each identical heavy sphere be $$M$$ and $$R$$ respectively. An object of mass $$m$$ be placed at the mid-point $$P$$ of the line joining their centres.
Force acting on the object placed at the mid-point,
$$F_1=F_2=\frac{G M m}{(5 R)^2}$$
The direction of forces are opposite, therefore net force acting on the object is zero.
To check the stability of the equilibrium, we displace the object through a small distance $$x$$ towards sphere $$A$$.
Now, force acting towards sphere $$A, F_1^{\prime}=\frac{G M m}{(5 R-x)^2}$$
Force acting towards sphere $$B, F_2^{\prime}=\frac{G M m}{(5 R+x)^2}$$
As $$F_1^{\prime}>F_2{ }^{\prime}$$, therefore a resultant force $$\left(F_1{ }^{\prime}-F_2{ }^{\prime}\right)$$ acts on the object towards sphere $$A$$, therefore object start to move towards sphere $$A$$ and hence equilibrium is unstable.