Problems on Critical Velocity and Period Of Satellite 02

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Example – 16:

  • A satellite revolves around a planet of mean density 104 kg/m3. If the radius of its orbit is only slightly greater than the radius of the planet, find the time of revolution of the satellite. G = 6.67 x 10-11S. I.  units.
  • Solutions:
  • Given: Density of material of planet = ρ = 104 kg/m3, G = 6.67 x 10-11 N m2/kg2 ;
  • To Find: Period of Satellite = T =?

The time period of a satellite orbiting around the earth is given by

If satellite orbiting very close to the earth  (i.e. h < < R) then h can be neglected. Then R + h = R



Ans: Time of revolution of the satellite is 1.044 hr

Example – 17:

  • A satellite makes ten revolutions per day around the earth. Find its distance from the earth assuming that the radius of the earth is 6400 km and g at the earth’s surface is 9.8 m/s2.
  • Solution:
  • Given: radius of earth = R = 6400 km = 6.4 x 106 m, g = 9.8 m/s2, Number of revolutions = 10 per day
  • To Find: radius of orbit of satellite = r =?

Ans: The radius of othe rbit of the satellite is 9125 km

Example – 18:

  • A satellite at a height of 2725 km makes ten revolutions around the earth per day. What is the number of revolution made per day by a satellite orbiting at a height of 560 km? Radius of earth = 6400 km.
  • Solution: 
  • Given: radius of orbit of earth =  R = 6400 km, height of satellite above the surface of earth in first case = h1 = 2725 km, height of satellite above the surface of earth in second case = h2 = 560 km, number of revolutions of satellite in first case = n1 = 10 per day
  • To Find: number of revolutions of satellite in second case = n2 = ?

r1 = R + h1 = 6400 + 2725 = 9125 km



r2 = R + h2 = 6400 +560 = 6960 km

Ans: The satellite will make 15 revolutions per day.

Example – 19:

  • What would be the speed of a satellite revolving in a circular orbit close to the earth’s surface? Given G = 6.67 x 10-11 S.I. units; density of earth’s matter = 5500 kg/m3 and radius of earth = 6400 km.  Also, find its period.
  • Solution:
  • Given: density of erth’s matter = ρ = 5500 kg/m3, radius of earth = R = 6400 km = 6.4 x 106 m, G = 6.67 x 10-11 N m2/kg2 ;
  • To Find: orbital velocity = vc = ?, period = T = ?

The time period of a satellite orbiting around the earth is given by



If satellite orbiting very close to the earth  (i.e. h < < R) then h can be neglected. Then R + h = R

The time period of a satellite orbiting around the earth is given by

T = 2πR/vc =  2 x 3.142 x 6400 /7.931 = 5071 s

T = 5071/3600 = 1.408 h



Ans: The speed of satellite is 7.931 km/s and time of revolution of the satellite is 1.408 h.

Example – 20:

  • Calculate the speed and period of revolution of a satellite orbiting at a height of 700 km above the earth’s surface. Assume the orbit to be circular. Take radius of the earth as 6400 km and g at the surface of the earth to be 9.8 m/s2.
  • Solution:
  • Given: height of satellite above the surface of earth = h = 700 km, Radius of earth = R = 6400 km, G = 6.67 x 10-11 N m2/kg2 ; g = 9.8 m/s2.
  • To Find: speed of satellite = vc = ?, Period = T = ?

r = R + h = 6400 + 700 = 7100 km = 7.1 x 106 m,

The speed of a satellite orbiting around the earth is given by

The time period of a satellite orbiting around the earth is given by



Ans: The speed of satellite is 7.519 km/s and period of revolution of the satellite is 5930 s

Example – 21:

  • A satellite is revolving around a planet in a circular orbit with a velocity of 8 km/s at a height where the acceleration due to gravity is 8 m/s2. How high is the satellite from the planet’s surface? Radius of planet = 6000 km.
  • Given: velocity of satellite vc = 8 km/s, R = 6400 km = 6.4 x 106 m,  acceleration due to gravity at height = gh = 8 m/s2.
  • To find: height of satellite above the surface = h = ?,

The critical velocity of a satellite orbiting around the earth is given by



Now, r = R + h

Thus h = r – R = 8000 – 6400 = 1600 km

Ans: The height of satellite above the surface of the earth is 1600 km.

Example – 22:

  • A satellite of mass 1750 kg is moving around the earth in an orbit at a height of 2000 km from the surface of the Earth. Find its angular momentum given G = 6.67 x 10-11 Nm2 kg-2 and mass of the  earth = 6 x  1024 kg.
  • Given: h = 2000 km, hence radius of orbit = r = R + h = 6400 + 2000 = 8400 km = 8.4 x 106 m, G = 6.67 x 10-11 Nm2 kg-2 ; mass of satellite = m = 1750 kg, mass of the earth = M = 6 x  1024 kg;
  • To find: Angular momentum = L= ?,

Ans : The angular momentum of the satellite is 1.015x 1014 kg m2 s-1.

Example – 23:

  • India’s most powerful rocket (PSLV) projected a remote sensing 850 kg satellite into an orbit 620 km above the earth’s surface. Assuming that the orbit is circular, find its speed and calculate the number of complete revolutions it makes around the earth in one day. g = 9.8 m/s2 and R = 6400 km.
  • Given: h = 620 km, radius of orbit of satellite = r = R + h = 6400 + 620 = 7020 km = 7.02 x 106 m, g = 9.8 m/s2, R = 6400 km  = 6.4 x 106 m
  • To find: speed of satellite = vc = ?, N = ?



 N = 14.81 revolutions per day

Ans: The speed of satellite = 7.561 km/s and number of revolutions of satellite per day = 14.81

Example – 24:

  • Two satellites X and Y are moving in circular orbits of radii r and 2r respectively around the same planet. What is the ratio of their critical velocities?
  • Given: rX = r, rY = 2r.
  • To find: vcX : vcY  =?

 Ans: The ratio of their critical velocities is 3 : 1



Example – 25:

  • A satellite going in a circular orbit of radius 4 x 104 km around the earth has a certain speed. If this satellite were to move around Mars with the same speed, what would be its orbital radius? The masses of earth and mars are in the ratio of 10:1 and their radii are in the ratio 2:1
  • Given: speeds are same vcE = vcM, radius of satellite around the earth = rE = 4 x 104 km, ME:MM = 10:1 and RE:RM = 2:1
  • To find: radius of the satellite around the mars =rM = ?

Ans: Radius of the satellite around Mars is 4000 km.

Example – 26:

  • The critical velocity of a satellite is 5 km/s. Find the height of satellite measured from the surface of the earth given G = 6.67 x 10-11 Nm2 kg-2; R=6400 km and M = 5.98 x 1024 kg
  • Given: critical velocity = vc = 5 km/s, 5 x 103 m, radius of earth = R = 6400 km = 6.4 x 106 m, G = 6.67 x 10-11 Nm2 kg-2;  mass of earthM = 5.98 x 1024 kg
  • To find: height of satellite above the surface of the earth = h = ?,

The critical velocity of a satellite orbiting around the earth is given by

Thus h = r – R = 15950 – 6400 = 9550 km

Ans: The height of satellite above the surface of the earth is 9550 km.



Example – 27:

  • A satellite is revolving around a planet in a circular orbit with a velocity of 6.8 km/s. Find the height of satellite from the planet’s surface and the period of its revolution. g = 9.8 m/s2 , R = 6400 km.
  • Given: velocity of satellite = vc = 6.8 km/s = 6.8 x 103 m/s, R = 6400 km = 6.4 x 106 m, g = 9.8 m/s2.
  • To find: height of satellite above the surface = h =?

The critical velocity of a satellite orbiting around the earth is given by

r = 8681 km

r = R + h

Thus h = r – R = 8681 – 6400 = 2281 km

Now,

Ans: The height of satellite above the surface of the earth is 2281 km

and period of the satellite is 8017 s

Example – 28:

  • A body is raised to a height of 1600 km above the surface of the earth and projected horizontally with a velocity 6 km/s. Will it revolve around the earth as satellite? 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.
  • Given:  radius of earth = 6.4 x 106 m = 6400 km, Mass of the earth = 6 x 1024 Kg, height of satellite above the surface =  h = 1600 km, horizontal velocity given to the satellite = v = 6 km/s, r = R + h = 6400 + 1600 = 8000 km, r = 8 x 106 m .
  • To find: the condition of orbiting of satellite = ?,

The critical velocity of a satellite orbiting around the earth is given by

  • The horizontal velocity given to satellite is 6 km/s which is less than the critical velocity 7.073 km/s. Hence the satellite will not revolve around the earth in circular orbit but will fall back on the earth in a parabolic path.

Example – 29:

  • A body is raised to a height equal to the radius of the earth above the surface of the earth and projected horizontally with a velocity 7 km/s. Will it revolve around the earth as satellite? If yes what is the nature of the orbit. Given G = 6.67 x 10-11 N m2/kg2; radius of earth = 6.4 x 106 m, Mass of the earth = 5.98 x 1024 Kg.
  • Given:  radius of earth = 6.4 x 106 m = 6400 km, Mass of the earth = 6 x 1024 Kg, height of satellite above the surface =  h = R, horizontal velocity given to the satellite = v = 7 km/s, r = R + R = 2R = 2 x 6400 = 12800 km, r = 12.8 x 106 m .
  • To find: the nature of the orbit of satellite = ?

The critical velocity of a satellite orbiting around the earth is given by

  • The horizontal velocity given to satellite is 7 km/s which is greater than the critical velocity 5.582 km/s and less than the escape velocity (11.2 km/s). Hence the satellite will revolve around the earth in elliptical orbit.

Example – 30:

  • With what velocity should a satellite be launched from a height of 600 km above the surface of the earth so as to move in a circular path?. Given: G = 6.67 x 10-11 N m2/kg2; radius of earth = 6.4 x 106 m, Mass of the earth = 5.98 x 1024 Kg.
  • Given: G = 6.67 x 10-11 N m2/kg2; radius of earth = R = 6.4 x 106 m = 6400 km, Mass of the earth = M = 5.98 x 1024 Kg, height of satellite above the surface of earth  = 600 km, raius of orbit = 6400 + 600 = 7000 km = .7 x 106 m
  • To find: critical velocity = vc = ?,

The critical velocity of a satellite orbiting around the earth is given by

Ans: The velocity of satellite for launching is 7.545 km/s.

Example – 31:

  • Two satellites orbiting round the earth have their initial speeds in the ratio 4 : 5. Compare their orbital radii.
  • Given: ratio of speeds vc1 : vc2  =  4 : 5
  • To find: r1 :  r = ?

The critical velocity of a satellite orbiting around the earth is given by

Ans : The ratio of their radii is 25 : 16

Example – 32:

  • A satellite is revolving around the earth in a circular orbit at a distance of 107 m from its centre. Find the speed of the satellite. Given G = 6.67 x 10-11 N m2/kg2, Mass of the earth = M =6 x 1024 Kg,
  • Given: radius of orbit  = 107 m, G = 6.67 x 10-11 N m2/kg2, Mass of the earth = M = 6 x 1024 Kg,
  • To find: speed of satellite = vc = ?,

The critical velocity of a satellite orbiting around the earth is given by

Ans: The speed of satellite is 6.326 km/s.

Example – 33:

  • The radii of the orbits of two satellites revolving around the earth are in the ratio 3:8. Compare their i) critical speed and ii) periods.
  • Given: ratio of radii of orbits r1 :  r =  3:8
  • To find: vc1 : vc2  =? , T1 : T2 = ?

Ans : The ratio of their critical velocities is 1.633:1

The ratio of their period is 0.2296:1

Example – 34:

  • The distances of two planets from the sun are 1013 m 1012 m respectively. Find the ratio of their periods and orbital speeds.
  • Given: r1 = 1013 m ,  r2 = 1012 m
  • To find: T1 : T2 = ?, vc1 : vc2 = ? ,

orbital speeds

Ans : The ratio of their period is 31.62:1

The ratio of their critical velocities is 0.3162:1

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

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