We can shield a charge from electric fields by putting it inside a hollow conductor. Can we shield a body from the gravitational influence of nearby matter by putting it inside a hollow sphere or by some other means?
A body cannot be shielded from the gravitational influence of nearby matter, because gravitational force between two point mass bodies is independent of the intervening medium between them.
It is due to the above reason, we cannot shield a body from the gravitational influence of nearby matter by putting it either inside a hollow sphere or by some other means.
An astronaut inside a small spaceship orbitting around the earth cannot detect gravity. If the space station orbitting around the earth has a large size, can he hope to detect gravity?
Inside a small spaceship orbitting around the earth, the value of acceleration due to gravity $g$, can be considered as constant and hence astronaut feels weightlessness. If the space station orbitting around the earth has a large size, such that variation in $g$ matters in that case astronaut inside the spaceship will experience gravitational force and hence can detect gravity. e.g., On the moon, due to larger size gravity can be detected.
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 $$0^{\circ}$$.