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If the height of a satellite from the earth is negligible in comparison to the radius of the earth R, the orbital velocity of the satellite is

gR/2

$\sqrt{g / R}$

$\sqrt{g R}$

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Explanation

(d) Orbital velocity near to earth = $\frac{v_{e}}{\sqrt{2}}$

$\Rightarrow v_{0} = \frac{\sqrt{2 g r}}{\sqrt{2}} = \sqrt{g r}$

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Choose the correct statement from the following : The radius of the orbit of a geostationary satellite depends upon -

Mass of the satellite, its time period and the gravitational constant

Mass of the satellite, mass of the earth and the gravitational constant

Mass of the earth, mass of the satellite, time period of the satellite and thegravitational constant

Mass of the earth, time period of the satellite and the gravitational constant

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Explanation

(d)

$T = 2 π \sqrt{\frac{r^{3}}{GM}} \Rightarrow r^{3} = \frac{G M T^{2}}{4 π^{2}} \Rightarrow r = {[\frac{G M T^{2}}{4 π^{2}}]}^{\frac{1}{3}}$

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If a body describes a circular motion under inverse square field, the time taken to complete one revolution T is related to the radius of the circular orbit as

$T \propto r$

$T \propto r^{2}$

$T^{2} \propto r^{3}$

$T \propto r^{4}$

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Explanation

(c)

$In a circular motion : F = \frac{{mv}^{2}}{r} = {mrω}^{2} In a inverse square field; F \propto \frac{1}{r^{2}} \Rightarrow {mrω}^{2} \propto \frac{1}{r^{2}} \Rightarrow ω^{2} \propto \frac{1}{r^{3}} \Rightarrow \frac{1}{T^{2}} \propto \frac{1}{r^{3}} \Rightarrow T^{2} \propto r^{3}$

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A planet moves around the sun. At a given point P, it is closest from the sun at a distance $d_{1}$ and has a speed $v_{1}$ . At another point Q, when it is farthest from the sun at a distance $d_{2}$ , its speed will be

$\frac{{d_{1}}^{2} v_{1}}{{d_{2}}^{2}}$

$\frac{d_{2} v_{1}}{d_{1}}$

$\frac{d_{1} v_{1}}{d_{2}}$

$\frac{{d_{2}}^{2} v_{1}}{{d_{1}}^{2}}$

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Explanation

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A satellite is moving around the earth with speed v in a circular orbit of radius r. If the orbit radius is decreased by 1%, its speed will

Increase by 1%

Increase by 0.5%

Decrease by 1%

Decrease by 0.5%

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Explanation

(b) $v \propto \frac{1}{\sqrt{r}} .$

% increase in speed = $\frac{1}{2}$ (% decrease in radius) $= \frac{1}{2} (1 %) = 0.5 %$

i.e. speed will increase by 0.5%

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Orbital velocity of an artificial satellite does not depend upon

Mass of the earth

Mass of the satellite

Radius of the earth

Acceleration due to gravity

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Explanation

(b) $v = \sqrt{\frac{G M}{r}}$

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The time period of a geostationary satellite is

24 hours

12 hours

365 days

One month

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Explanation

(a) $ω_{s} = ω_{e} \Rightarrow \frac{2 π}{T_{s}} = \frac{2 π}{T_{e}} o r, T_{s} = T_{e} (24 h o u r s)$

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Two planets move around the sun. The periodic times and the mean radii of the orbits are $T_{1}, T_{2}$ and $r_{1}, r_{2}$ respectively. The ratio $T_{1} / T_{2}$ is equal to

${(r_{1} / r_{2})}^{1 / 2}$

$r_{1} / r_{2}$

${(r_{1} / r_{2})}^{2}$

${(r_{1} / r_{2})}^{3 / 2}$

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Explanation

(d) $T^{2} \propto r^{3} \Rightarrow \frac{T_{1}}{T_{2}} = {(\frac{r_{1}}{r_{2}})}^{3 / 2}$

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Orbital velocity of earth's satellite near the surface is 7 km/s. When the radius of the orbit is 4 times than that of earth's radius, then orbital velocity in that orbit is

3.5 km/s

7 km/s

72 km/s

14 km/s

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Explanation

(a) $v \propto \frac{1}{\sqrt{r}} .$ If orbital radius becomes 4 times then orbital velocity will

become half. i.e. $\frac{7}{2} = 3.5 k m / s$

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The largest and the shortest distance of the earth from the sun are $r_{1}$ and $r_{2}$ , its distance from the sun when it is at the perpendicular to the major axis of the orbit drawn from the sun

$\frac{r_{1} + r_{2}}{4}$

$\frac{r_{1} r_{2}}{r_{1} + r_{2}}$

$\frac{2 r_{1} r_{2}}{r_{1} + r_{2}}$

$\frac{r_{1} + r_{2}}{3}$

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Explanation

$r_{1} = (1 + e) a a n d r_{2} = (1 - e) a \Rightarrow a = \frac{r_{1} + r_{2}}{2} a n d r_{1} r_{2} = (1 - e^{2}) a^{2}$

where a = semi major axis

b = semi minor axis

e = eccentricity

Now required distance = semi latus rectum= $\frac{b^{2}}{a}$

= $\frac{a^{2} (1 - e^{2})}{a} = \frac{(r_{1} r_{2})}{(r_{1} + r_{2}) / 2} = \frac{2 r_{1} r_{2}}{r_{1} + r_{2}}$

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