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 $$0^{\circ}$$.
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.