Physics 5 Marks Questions

1. No, from the formula $\upsilon_\text{e}=\sqrt{\frac{2\text{GM}}{\text{R}}},$ it is clear that escape velocity does not depend on the mass of the body.
2. The escape velocity depends upon the value of gravitational potential at the point from where the body is projected. The gravitational potential energy of body $\text{E}=-\frac{\text{GMm}}{\text{R}}$ is slightly different at different points ($\because$ the earth is not a perfect sphere and hence R is different ai different points). Because of this escape velocity depend slightly on the latitude of the place from where the body is projected.
3. The escape velocity of a body does not depend upon its direction of projection.
4. Since the gravitational potential energy at a point at the height h from the earth surface is $\frac{\text{GMs}}{\text{(R+h)}},$ the escape velocity will be different for different values of h.

$=-\frac{\big(\frac{1}{2}\big)\text{GM}_\text{e}\text{m}}{(\text{R}_\text{e}+\text{h})}$

$=\frac{\big(\frac{1}{2}\big)\text{GM}_\text{e}\text{m}}{(\text{R}_\text{e}+\text{h})}$

$=\frac{\big(\frac{1}{2}\big)\times6.67\times10^{-11}\times6\times10^{24}\times200}{(6.4\times10^6+0.4\times10^6)}$

$=5.9\times10^9\text{ j.}$

$=\bigg(\frac{4\times(3.14)^2\times(4.73)^2\times10^{60}}{6.67\times10^{-11}\times5\times10^{41}}\bigg)^\frac{1}{2}=\bigg(\frac{39.48\times105.82\times10^{30}}{33.35}\bigg)^{\frac{1}{2}}$

$=(125.27\times10^{30})^\frac{1}{2}=1.12\times10^{16}\text{ s}$

$1\text{ year}=365\times324\times60\times60\text{s}$

$1\text{s}=\frac{1}{365\times24\times60\times60}\text{ year}$

$\therefore\ 1.12\times10^{16}\text{ s}=\frac{1.12\times10^{16}}{365\times24\times60\times60}$

$=3.55\times10^{8}\ \text{year}$

$\frac{1}{2}\text{mv}_\text{p}^2-\frac{1}{2}\text{mv}_\text{esc}^2=\frac{1}{2}\text{mv}_\text{f}^2$

$\text{v}_\text{f}=\sqrt{\text{v}_\text{p}^2-\text{v}_\text{esc}^2}$

$=\sqrt{(3\text{v}_\text{esc})^2-(\text{v}_\text{esc})^2}$

$=\sqrt{8}\text{v}_\text{esc}$

$=\sqrt{8}\times11.2=31.68\text{km/s}$

$\omega=$ Angular speed $=2\pi\text{v}$

$=\Big(\frac{1}{2}\Big)\text{mv}^2+\Big(\frac{-\text{GM}_\text{e}\text{m}}{\text{R}_\text{e}}\Big)$

$\text{g}'=\frac{\text{g}}{\big[1+\big(\frac{\text{h}}{\text{R}_\text{e}}\big)\big]^2}$

For $\text{h}=\frac{\text{R}_\text{e}}{2}$

$\text{g}'=\frac{\text{g}}{\big[1+\big(\frac{\text{R}_\text{e}}{2\text{R}_\text{e}}\big)\big]^2}$

$=\frac{\text{g}}{\big[1+\big(\frac{1}{2}\big)\big]^2}=\Big(\frac{4}{9}\Big)\text{g}$

Orbital radius of Io, R_Io = 4.22 × 10⁸m

Satellite Io is revolving around the Jupiter

Mass of the satellite is given as:

$\text{M}_\text{j}=\frac{(4\pi^2\text{R}_\text{Io}^3)}{(\text{GT}_\text{Io}^2)}$

Where,

M_J = Mass of Jupiter

G = Universal gravitational constant

Orbital period of the Earth, T_e = 365.25 days = 365.25 × 24 × 60 × 60s

Orbital radius of the Earth,

R_e = 1AU = 1.496 × 10¹¹m

Mass of the Sun $\text{M}_\text{s}=\frac{(4\pi^2\text{R}_\text{e}^3)}{(\text{GT}_\text{e}^2)}\ ....(\text{ii})$

Therefore, $\Big(\frac{\text{M}_\text{s}}{\text{M}_\text{j}}\Big)=\frac{(4\pi^2\text{R}_\text{e}^3)}{(\text{GT}_\text{e}^2)}\times\frac{(\text{GT}_\text{Io}^2)}{(4\pi^2\text{R}_\text{Io}^3)}$

= ((1.769 × 24 × 60 × 60)/ (365.25 x 24 x 60 x60))² × ((1.496 x 10¹¹)/ (4.22 × 10⁸))³

= 1045.04

Therefore, (M_s/ M_J) ≈ 1000

M_s ≈ 1000M_J

This shows the mass of Jupiter is (1/1000^th) the mass of the sun.

1. The minimum speed required for an object to reach infinity (i.e., to get escape from earth) is called escape velocity.

2. $\frac{1}{2}\text{m}\Big(\text{v}_{\text{i}}^2\Big)_{\text{e}}=\frac{\text{GmM}_{\text{p}}}{\text{h+R}_{\text{p}}}$

$\begin{bmatrix}\text{where M}_{\text{p}}=\text{Mass of planet}\\\ \ \ \ \ \ \ \ \ \ \text{R}_{\text{p}}=\text{Radius of planet}\end{bmatrix}$

If the object is thrown from surface of a planet h = 0, we get

$(\text{v}_{\text{i}})_{\text{e}}=\sqrt{\frac{2\text{GM}_{\text{p}}}{\text{R}_{\text{p}}}}$

$\text{but }\text{g}=\frac{\text{GM}_{\text{p}}}{\text{R}_{\text{p}}}\text{ we get}$

$(\text{v}_{\text{i}})_{\text{e}}=\sqrt{2\text{gR}_{\text{p}}}$

3. Depends on location as 'g' varies with location, as most of the celestial bodies are not perfectly spherical.

1. The particles will always remain diametrically opposite so that the force on each particle will be directed along the radius. Consider the motion of one of the particles. The force on the particle is given by

$\text{F}=\frac{\text{Gm}^2}{4\text{R}^2}$

$\therefore\frac{\text{Gm}^2}{4\text{R}^2}=\frac{\text{mv}^2}{\text{R}}$

$\Rightarrow\text{v}=\sqrt{\frac{\text{Gm}}{4\text{R}}}$

2. Binding energy is the minimum energy required to free a satellite from the gravitational attraction.

$\text{T.E.}=\text{P.E.}+\text{K.E.}=-\frac{\text{GMm}}{\text{R}}+\frac{1}{2}\text{mv}^2$

$\text{T.E.}=-\frac{\text{GMm}}{\text{R}}+\frac{\text{mGM}}{\text{R}}=\frac{-\text{GMm}}{\text{2R}}$

Binding energy (B.E.) = -(T.E.) $=\frac{\text{GMm}}{2\text{R}}$

r_p = a(1 - e) and r_a = a(1 + e)...(i)

As the earth orbits the sun, the angular momentum is conserved and a real velocity is constant.

Let M be the mass of the earth,

v_p = velocity of the earth at perihelion (perigee).

v_a = velocity of the earth at aphelion or apogee.

$\omega_\text{p}$ = angular velocity of the earth at perihelion.

$\omega_\text{a}$ = angular velocity of the earth at aphelion.

Angular momentum and areal velocity are constant as the earth orbits the sun.

$\text{r}^2_\text{p}\omega_\text{p}=\text{r}^2_\text{a}\omega_\text{a}...(\text{ii})$

From (i) and (ii), we get

$\frac{\omega_\text{p}}{\omega_\text{a}}=\Big(\frac{1+\text{e}}{1-\text{e}}\Big)^2\text{e}=0.0167$

$\therefore\ \frac{{\omega_\text{p}}}{{\omega_\text{a}}}=1.0691$

Let $\omega$ be the mean angular speed corresponds to mean solar day.

$\therefore\ \Big(\frac{{\omega_\text{p}}}{\omega}\Big)\Big(\frac{\omega}{{\omega_\text{a}}}\Big)=1.0691$

$\omega^2={\omega_\text{p}}{\omega_\text{a}}$

$\therefore\ \frac{{\omega_\text{p}}}{\omega}=\frac{\omega}{{\omega_\text{a}}}=1.034$

If mean angular velocity co corresponds to 1° per day, then _p = 1.034° per day and _a = 0.967° per day.

Since, 361° = 24 mean solar day we get (360 + 1.034) which corresponds to 24 h, 8.14" (8.1" longer) and 360.967°, corresponds to 23 h 59 min 52" (7.9" smaller).

This does not explain the actual variation of the length of the day during the year.

According to the diagram,

r_p = radius of perigee = 2R

r_a = radius of apogee = 6R

a = semi - major axis of the ellipse

Hence, we can write

r_a = a(1 + e) = 6R

P_p = a(1 - e) = 2R

$\frac{\text{a(1+e)}}{\text{a(1}-\text{e})}=\frac{6\text{R}}{2\text{R}}=3$

By solving, we get eccentricity $\text{e}=\frac{1}{2}$

If v_a and v_p are the velocities of the satellite (of mass m) at aphelion and perihelion respectively, then by conservation of angular momentum

$\therefore\ \text{L}_\text{at perigee}=\text{L}_\text{at apogee}$

$\text{mv}_\text{p}\text{r}_\text{p}=\text{mv}_\text{a}\text{r}_\text{a}$

$\therefore\ \frac{\text{v}_\text{a}}{\text{v}_\text{p}}=\frac{\text{r}_\text{p}}{\text{r}_\text{a}}=\frac{1}{3}$

Applying conservation of energy,

Energy at perigee = Energy at apogee

where M is the mass of the earth

$\therefore\ \text{v}_\text{p}^2\Big(1-\frac{1}{9}\Big)=-2\text{GM}\Big(\frac{1}{\text{r}_\text{a}}-\frac{1}{\text{r}_\text{p}}\Big)$

$=2\text{GM}\Big(\frac{1}{\text{r}_\text{p}}-\frac{1}{\text{r}_\text{a}}\Big)$ $(\text{By putting v}_\text{a}=\frac{\text{v}_\text{p}}{3})$

$\text{v}_\text{p}=\frac{\Big[2\text{GM}\big(\frac{1}{\text{r}_\text{p}}-\frac{1}{\text{r}_\text{a}}\big)\Big]^{\frac{1}{2}}}{\Big[1-\big(\frac{\text{v}_\text{a}}{\text{v}_\text{p}}\big)^2\Big]^{\frac{1}{2}}}$

$=\Bigg[\frac{\frac{2\text{GM}}{\text{R}}\Big(\frac{1}{2}-\frac{1}{6}\Big)}{\Big(1-\frac{1}{9}\Big)}\Bigg]^{\frac{1}{2}}=\Big(\frac{\frac{2}{3}}{\frac{8}{9}}\frac{\text{GM}}{\text{R}}\Big)^{\frac{1}{2}}=\sqrt{\frac{3}{4}\frac{\text{GM}}{\text{R}}}=6.85\text{km/ s}$

$\text{v}_\text{p}=6.85\text{km/ s,}\text{ v}_\text{a}=2.28\text{km/ s,}$

For circular orbit of radius r,

v_c = orbital velocity $=\sqrt{\frac{\text{GM}}{\text{r}}}$

For r = 6R, v_c $=\sqrt{\frac{\text{GM}}{\text{6R}}}$ = 3.23km/ s

$=\frac{\text{R}^2\omega}{\text{R}^2\Big[1-\big(\frac{1}{\text{n}}\big)\Big]^2}$

$=-\frac{3}{4}\frac{\text{GMm}}{\text{R}}$

$\Rightarrow\frac{1}{\text{R}+\text{h}}=\frac{3}{4\text{R}}\text{ or }3\text{R}+3\text{h}=4\text{R}$

$\Rightarrow3\text{h}=4\text{R}-3\text{R}=\text{R}$

$\text{or }\text{h}=\frac{\text{R}}{3}$

$=-\frac{\big(\frac{1}{2}\big)\text{GM}_\text{e}\text{m}}{(\text{R}_\text{e}+\text{h})}$

$=\frac{\big(\frac{1}{2}\big)\text{GM}_\text{e}\text{m}}{(\text{R}_\text{e}+\text{h})}$

$=\frac{\big(\frac{1}{2}\big)\times6.67\times10^{-11}\times6\times10^{24}\times200}{(6.4\times10^6+0.4\times10^6)}$

$=5.9\times10^9\text{ j.}$

$\text{g}'=\frac{\text{g}}{\big[1+\big(\frac{\text{h}}{\text{R}_\text{e}}\big)\big]^2}$

For $\text{h}=\frac{\text{R}_\text{e}}{2}$

$\text{g}'=\frac{\text{g}}{\big[1+\big(\frac{\text{R}_\text{e}}{2\text{R}_\text{e}}\big)\big]^2}$

$=\frac{\text{g}}{\big[1+\big(\frac{1}{2}\big)\big]^2}=\Big(\frac{4}{9}\Big)\text{g}$

1. The gravitational force 'mg' towards centre O.
2. The centrifugal force trying to lift the mass away. The net force is given by,

$\text{mg}_{\theta}=\sqrt{(\text{mg})^2+(\text{mR}\cos\theta\ \omega^2)^2+2(\text{mg})\$\text{mR}\cos\theta\ \omega^2)\cos(180^\circ-\theta)}$

The radius is $\text{R}\cos\theta$ since the centrifugal force is due to the rotation about the axis of the earth.

$\text{mg}_{\theta}=\text{mg}\Bigg(1+\Big(\frac{\text{R}\omega^2\cos^2\theta}{\text{g}}\Big)^2-2\frac{\text{R}\omega^2}{\text{g}}\cos^2\theta\Bigg)^{\frac{1}{2}}$

$\therefore\text{g}_{\theta}\Big(1-\frac{2\text{R}\omega^2}{\text{g}}\cos^2\theta\Big)^{\frac{1}{2}}$

Since $\Big(\frac{\text{R}\omega^2}{\text{g}}\Big)^2$ is negligibly small.

$\text{g}_{\theta}=\text{g}\Big(1-\frac{\text{R}\omega^2}{\text{g}}\cos^2\theta\Big)$

 $\therefore\text{P.E}(\text{U})=\frac{-\text{GMm}}{\text{r}}$

1. All the planets move in elliptical path with sun at their foci.
2. The line joining the sun and the planet sweeps out equal areas in equal intervals of time.
3. The square of the time period of revolution is proportional to the cube of the semi-major axis of the elliptical orbit $\text{T}^2\propto\text{a}^3.$

If r₁ and r₂ are the shortest and the longest distances of the planet from the sun, the semi-major axis is given by $\Big(\frac{\text{r}_1+\text{r}_2}{2}\Big)$

Ist law is proved from angular momentum conservation in a circular path.

Since L = 2m × aerial velocity, for equal time, area swept will be same. So 2^nd law is proved.

From $\text{F}\propto\frac{1}{\text{r}^2},\frac{\text{mv}^2}{\text{r}}=\text{F}\text{ and }\text{T}=\frac{2\pi\text{r}}{\text{v}}$ one can prove

$\text{T}^2\propto\text{a}^3$ thereby proving 3rd law

1. Distance r is, V $=\frac{-\text{GM}}{\text{r}}$

Potential at a point A (r_a) $=-\frac{\text{GM}}{\text{r}_{\text{a}}}$

Potential at a point B (r_b) $=-\frac{\text{GM}}{\text{r}_{\text{b}}}$

$\therefore$ Difference in Potential between the points

$=-\text{GM}\Big(\frac{1}{\text{r}_{\text{a}}}-\frac{1}{\text{r}_{\text{b}}}\Big)$

2. Since work done against gravitational force is (a) independent of path and dependent only on the initial and final points and (b) the work done in a closed path is zero, it is a conservative force.