Tag: moon and stars in sky

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Questions Related to moon and stars in sky

Multiple choice evs artificial satellite types of artificial satellites geostationary orbits moon and stars in sky

Satellite is revolving around the earth. If it's radius of orbit is increased to $4$ times the radius of geostationary satellite, what will become its time period ?

  1. $8\ days$

  2. $4\ days$

  3. $2\ days$

  4. $16\ days$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given,
$r _1=r$ 
$r _2=4r$
$T _1=1day$
The time period of the geostationary satellite is given by
$T=2\pi \sqrt{\dfrac{r^3}{Gm _E}}$. . . . . .(1)
where, $G=$ gravitational constant 
From equation (1),
$T\propto \sqrt{r^3}$
$\dfrac{T _2}{T _1}=\sqrt{\dfrac{r _2^3}{r _1^3}}$
$T _2=T _1\sqrt{\dfrac{4r\times 4r\times 4r}{r^3}}=8T _1 days$
$T _2=8\times 1=8days$
The correct option is A.

Multiple choice evs artificial satellite types of artificial satellites geostationary orbits moon and stars in sky

The distance between the centre of the earth and moon is 384000 km. If the mass of the earth is $6 \times 10^{24} kg$ and $G=6.66\times 10^{-11}$ units,the speed of the moon is nearly

  1. 1 km/s

  2. 4 km/s

  3. 8 km/s

  4. 11.2 km/s
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Orbital speed v = sqrt(GM/r). G = 6.67 * 10^-11, M = 6 * 10^24, r = 3.84 * 10^8 m. v = sqrt(6.67 * 10^-11 * 6 * 10^24 / 3.84 * 10^8) = sqrt(40.02 * 10^13 / 3.84 * 10^8) = sqrt(10.4 * 10^5) approx 1020 m/s = 1 km/s.

Multiple choice evs artificial satellite types of artificial satellites geostationary orbits moon and stars in sky

A satellite orbiting close to the earth's surface will escape if 

  1. its speed is increased by 41.4%

  2. its KE is made 1.5 times the original value

  3. its original speed is increased $\sqrt{1.5}$ times

  4. its stops moving in the orbit
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Escape velocity v_e = sqrt(2) * v_o, where v_o is orbital velocity. v_e = 1.414 * v_o. To escape, the speed must increase by 0.414 * v_o, which is 41.4%.

Multiple choice evs artificial satellite types of artificial satellites geostationary orbits moon and stars in sky

A particle is projected with a velocity $\sqrt{\dfrac{4gR}{3}}$ vertically upward from the surface of the earth. R is the radius of the earth & g being the acceleration due to gravity on the surface of the earth. The velocity of the particle when it is at half the maximum height reached by it is:

  1. $\sqrt{\dfrac{gR}{2}}$

  2. $\sqrt{\dfrac{gR}{3}}$

  3. $\sqrt{gR}$

  4. $\sqrt{\dfrac{2gR}{3}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using energy conservation: (1/2)mv^2 - GMm/R = -GMm/(R+h). Given v = sqrt(4gR/3) and g = GM/R^2, we find max height h. Then find velocity at h/2 using energy conservation again.

Multiple choice evs artificial satellite types of artificial satellites geostationary orbits moon and stars in sky

The relay satellite transmits the television programme continuously from one part to another because its : 

  1. Period is greater than the period of rotation of the earth about its axis

  2. Period is less than the period of rotation of the earth about its axis

  3. Period is equal to the period of rotation of the earth about its axis

  4. Mass is less than the mass of earth
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Relay satellite has to forces on the are unstartly suppose if the satellite is stable in space of a certain position, then as the earth rotate satellite can show once one part and later another part which is absorb movement. Therefore the satellite has to revolve around the earth with same angular speed. Therefore the time period of satellite revolution should be same as time period of the rotation of earth and rotation of the satellite should be about axis of earth's rotation. So its period is equal to period of rotation of the earth about its axis.
Multiple choice evs artificial satellite types of artificial satellites geostationary orbits moon and stars in sky

If mass of earth is $5.98\times 10^{24}$ kg and earth moon distance is $3.8\times 10^5$ km, the orbital period of moon, in days is

  1. 27 days

  2. 2.7 days

  3. 81 days

  4. 8.1 days
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$M=5.98 \times 10^{24}kg$


$R=3.8\times 10^8km$


The orbital time period of moon,

$T=2\pi \sqrt{\dfrac{R^3}{GM}}$

$T=2\times 3.14\sqrt{\dfrac{(3.8)^3\times 10^{24}}{6.67\times 10^{-11}\times 5.48 \times 10^{24}}}$

$T=23.2427 \times 10^5 sec$

In days,

Time period, $T=\dfrac{23.2427\times 10^5}{60\times 60\times 24}day$

$T=27\ days$

The correct option is A.

Multiple choice evs artificial satellite types of artificial satellites geostationary orbits moon and stars in sky

If the angular velocity of a planet about its own axis is halved, the distance of geostationary satellite of this planet from the cent of the planet will become :

  1. $(2)^{1/3}$ times

  2. $(2)^{3/2}$ times

  3. $(2)^{2/3}$ times

  4. 4 times
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Lets consider, 

$\omega _1=\omega $ angular velocity

$\omega _2=\dfrac{\omega }{2}$

$r=$ distance between geostationary satellites.

The Time period, $T\propto r^{3/2}$

$\dfrac{2\pi}{\omega }\propto r^{3/2}$

$\omega \propto r^{-3/2}$. . . . .(1)

$\dfrac{\omega _2}{\omega _1}\propto(\dfrac{r _2}{r _1})^{-3/2}$

$\dfrac{1}{2}\propto (\dfrac{r _2}{r _1})^{-3/2}$

$r _2=(2)^{3/2} r _1$

The correct option is B.
Multiple choice evs artificial satellite types of artificial satellites geostationary orbits moon and stars in sky

A satellite has to revolve round the earth in a circular orbit of radius 8 x $10^3$km. The velocity of projection of the satellite in this orbit will be -

  1. 16 km/sec

  2. 8 km/sec

  3. 3 km/sec

  4. 7.08 km/sec
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Orbital velocity v = sqrt(GM/r). G = 6.67 * 10^-11, M = 6 * 10^24, r = 8 * 10^6 m. v = sqrt(6.67 * 10^-11 * 6 * 10^24 / 8 * 10^6) = sqrt(40 * 10^13 / 8 * 10^6) = sqrt(5 * 10^7) approx 7071 m/s = 7.07 km/s.