Problems on Critical Velocity and Period Of Satellite 01

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Example – 1:

  • The mean angular velocity of the earth around the sun is 1° per day. The distance from the sun to the earth is 1.5 x 10km. Determine the mass of the sun. G = 6.67 x 10-11 S.I. units.
  • Solution:
  • Given: Angular frequency = ω = 1° per day, radius of orbit = r = 1.5 x 108 km = 1.5 x 1011 m, G = 6.67 x 10-11 S.I. units.
  • To find: mass of sun = M =?

∴ T = 2 x 24 x 60 x 60 x 180    s

∴  M = 2.062 x 1030 kg



Ans: Mass of the sun is 2.062 x 1030 kg

  • Calculator Calculation:

(4 x 3.14^2 x 1.5^3x 10^23)÷(6.67 x 2^2 x 24^2 x 60^4 x 180 ^2)

  • Log Calculation:

Example – 2:

  • Find the mass of the sun given that the radius of earth’s orbit is 1.5 x 108km. G = 6.67 x 10-11 S. I. units and the period of earth’s revolution around the sun is 365 days.
  • Solution:
  • Given: Period of revolution = T= 365 days = 365 x 24 x 60 x 60 s, Radius of orbit = r = 1.5 x 108 km = 1.5 x 1011 m, G = 6.67 x 10-11 S.I. units.
  • To find: mass of sun = M =?



∴  M = 2.062 x 1030 kg

Ans: Mass of sun is 2.062 x 1030 kg

Example – 3:

  • From the following data, calculate the period of revolution of the moon around the earth : Radius of earth = 6400 km; Distance of moon from the earth = 3.84 x 105 km; g = 9.8 m/s2.
  • Solution:
  • Given: radius of earth = R = 6400 km = 6.4 x 108 m, radius of orbit of moon = r = 3.84 x 105 km = 3.84 x 108 m,  g = 9.8 m/s2.
  • To find: Period of revolution of moon = T = ?

Ans: Period of the moon around the earth is 27.3 days

  • Calculator Calculation:

2 x 3.142 x ((3.84^2 x 10^24)÷(6.4^2 x 10^12 x 9.8))^(1÷2)



  • Log Calculation:

Example – 4:

  • If the moon revolves around the earth once in 27 days and 7 hours in an orbit which is 60 times the earth’s radius, find g at the earth’s surface. R= 6400 km.
  • Given: radius of earth = R = 6400 km = 6.4 x 106 m, radius of orbit of moon = r = 60 R , Period of revolution of ,moon = 27 days 7 hours =  = 27 x 24 x 60 x 60+ 7 x 60 x 60 =6556060 s,
  • To find: acceleration due to gravity = g = ?

Ans: g on the Earth’s surface = 9.81 m/s2

Example – 5:

  • Find the radius of moon’s orbit around the earth assuming the orbit to be circular. Period of revolution of the moon around the earth = 27.3 days, g at the earth’s surface = 9.8 m/s2. Radius of earth = 6400 km.
  • Given: Radius of earth = R = 6400 km = 6.4 x 106 m, Time period = T = 27.3 days = 27.3 x 24 x 60 x 60, g = 9.8 m/s6.
  • To find: radius of the orbit = r =?



Ans: The radius of moon’s orbit is 3.841 x 10km

  • Calculator Calculation:

((27.3^2 x 24^2 x 60^4 x 6.4^2 x 10^12 x 9.8) ÷ (4 x 3.14^2))^ (1÷3)

  • Log Calculation:

Example – 6:

  • Assuming that the moon describes a circular orbit of radius R about the earth in 27 days and that Titan describes a circular orbit of radius 3.2 R about Saturn in 16 days, compare the mass of Saturn and the earth.
  • Solution:
  • Given: Period of Moon = TM= 27 days, radius of orbit of moon = rM = R, time period of titan = TT =16 days, radius of orbit of titan =  rT = 3.2 R,
  • To Find:   Ratio of mass of saturn to the earth = MS /ME=?



Ans: The ratio of the the mass of Saturn to the earth is 93.3 :1

Example – 8:

  • Compare the critical speeds of two satellites if the ratio of their periods is 8 : 1.
  • Solution:
  • Given: T1/T2 = 8/1
  • To Find: Ratio of critical velocity = vc1/vc2 =?,

From equations (1) and (2)



Ans: The ratio of their critical velocities is 1:2

Example – 9:

  • An artificial satellite is revolving around the earth in circular orbit at a height of 1000 km at a speed of 7364 m/s. Find its period of revolution if R = 6400 km
  • Solution:
  • Given: Radius of earth = R = 6400 km = 6.4 x 106 m, height of satellite above the surface of the earth = h = = 1000 km = 1.0  x 106 m, radius of orbit of satellite = r = R + h = 6.4 x 106  +  1.0 x 106 m = 7.4 x 10 m, critical velocity = vc =7364 m/s..
  • To Find: T = ?

 

Ans: Period of revolution of the satellite is 1.75 hr.

Example – 10:

  • Moon takes 27 days to complete one revolution around the earth. calculate its linear velocity. If the distance between the earth and the moon is 3.8 x  10km.
  • Solution:
  • Given: Radius of orbit of moon = r = 3.8 x 105 km =3.8 x 108 m, Period of moon = T = 27 days = 27 24  60  60 s.
  • To find: critical velocity = vc =?

 



Ans:  The linear velocity of the moon is 1.023 km/s

Example – 11:

  • A  communication satellite is at a height of 36400 km from the surface of the earth. What will be its new period when it is brought down to the height of 20000 km from the surface of the earth. The radius of the earth is 6400 km.
  • Solution:
  • Given: Initial radius of orbit = r1 = 36400 + 6400 = 42800 km, Final radius of orbit = r2=   6400 + 20000 = 26400 km , Initial time period of satellite = T1 =24 hr,
  • To Find:   New time period of satellite = T2 =?

By Keppler’s law

∴ T2 = 11.62 h



Ans: The new period of satellite 11.62 hr

  • Log calculation:

Example – 12:

  • Venus is orbiting around the sun in 225 days. Calculate the orbital radius and speed of the planet.  Mass of sun is   2 x 1030 kg.                   G = 6.67 x 10-11 S.I. units.
  • Solution:
  • Given: Mass of sun = M = 2 x 1030 kg, Period of venus = T = 225 days = 225 x 24 x 60 x 60 s, G = 6.67 x 10-11 S.I. units.
  • To Find: radius of orbit = r =? orbital velocity = vc =?

Ans: Orbital radius of Venus = 1.085 x 1011 m and its orbital speed =  3.506 x 104 m/s

Example – 13: 

  • An observer situated at the equator finds that a satellite is always overhead. What must be its distance from the centre of the earth? Given g at earth’s surface = 9.81 m/s2 ; radius of earth = 6.4 x 106 m. What is the KE of such a satellite w.r.t. an observer on earth? What must be its height above the surface of the earth?
  • Solution:
  • Given: Period of Earth = T = 24 hr = 24 x 60 x 60 s, Radius of the cearth = R = 6.4 x 106 m, g = 9.8 m/s2.
  • To Find: radius of the orbit = r =?



Now, r = R + h

Hence, h = r – R = 42350 – 6400 = 35950 km

Ans: Radius of the orbit of the satellite is 42350  km. The height of satellite above the surface of the earth is 35950 km.

As the satellite is stationary w.r.t. observer, the kinetic energy of satellite w.r.t. the observer is zero.

Example – 14:

  • A satellite is revolving around the earth in a circular orbit in the equatorial plane at a height of 35850 km. Find its period of revolution. What is the possible use of such a satellite? In what direction is such a satellite projected and why must it be in the equatorial plane? Given g = 9.81 m/s2 ; Radius of earth 6.37 x 106 m
  • Solution:
  • Given: Radius of orbit of the earth = R = 6.37 x 106 m, height of satellite above the surface of the earth = h = 35850 km = 35850 x 103 m = 35.85 x 106 m, G = 6.67  10-11 N m2/kg2,  g = 9.8 m/s2.
  • To Find: T = ?

r = R + h = 6.37 x 106 m +  35.85 x 106 m = 42.22 x 106 m

Ans: Period of the satellite is 24 hr.

  • Such satellite is called geosynchronous satellite and is used for communication, broadcasting and weather forecasting. Such satellite moves in the same direction as that of the rotation of the earth and its orbit is in the equatorial plane.If this satellite is not in the equatorial plane, it will appear to move up and down of the equatorial plane and thus it will not be stationary w.r.t. observer on the Earth.

Example – 15:

  • Calculate the height of communication satellite above the surface of the earth? Given G = 6.67 x 10-11 N m2/kg2 ; radius of earth  = 6.4 x 106 m, Mass of the earth = 6 x 1024 Kg.
  • Solution:
  • Given: Period of satellite = T = 24 hr = 24 x 60 x 60 s, G = 6.67 x 10-11 N m2/kg2 ; radius of earth = R  = 6.4 x 106 m, mass of earth = M = 6 x 1024 Kg
  • To Find: height of satellite above the surface of the earth = h = ?

Critical Velocity

Now, r = R + h

Hence, h = r – R = 42.32 x 106 – 6.4 x 10= 35.92 x 106 m

h = 35920  x 10 m = 35920 km

Ans: The height of satellite above the surface of the earth is 35920 km.

Science > Physics > Gravitation > You are Here
Physics Chemistry Biology Mathematics

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