Q 1 :    

Two satellite A and B go round a planet in circular orbits having radii 4R and R respectively. If the speed of A is 3v, the speed of B will be           [2024]

  • 6v

     

  • 12v

     

  • 43v

     

  • 3v

     

(1)

 



Q 2 :    

A satellite revolving around a planet in stationary orbit has time period 6 hours. The mass of planet is one-fourth the mass of earth. The radius orbit of planet is

(Given = Radius of geo-stationary orbit for earth is 4.2×104 km)          [2024]

  • 1.05×104 km

     

  • 1.4×104 km

     

  • 1.68×105 km

     

  • 8.4×104 km

     

(1)   

        By Kepler's law, T2=4π2R3GMT2R3M

         (T1T2)2=(R1R2)3(M2M1)

         (246)2=(4.2×104R2)3×(ME4ME)

         42×4=(4.2×104R2)3R2=1.05×104 km

 



Q 3 :    

Correct formula for height of a satellite from Earth's surface is _____.     [2024]

  • (T2R2g4π2)1/3-R

     

  • (T2R2g4π2)1/2-R

     

  • (T2R24π2g)1/3-R

     

  • (T2R2g4π2)-1/3+R

     

(1)

mv2(R+h)=GMm(R+h)2v=GMR+h

Now T=2π(R+h)GMR+h

T=2π(R+h)3/2GM=2π(R+h)3/2Rg

T2R2g=4π2(R+h3)

(T2R2g4π2)1/3-R=h



Q 4 :    

A satellite of mass M2 is revolving around earth in a circular orbit at a height of R3 from earth surface. The angular momentum of the satellite is MGMRx. The value of x is ________, where M and R are the mass and radius of earth, respectively. (G is the gravitational constant)          [2025]



(3)

(i) If earth is assumed to be stationary

Orbital velocity v0=GM4R/3=3GM4R

Angular momentum of satellite

L=M2v04R3

L=M2·3GM4R·4R3=MGMR3

 x=3



Q 5 :    

A satellite of mass 1000 kg is launched to revolve around the earth in an orbit at a height of 270 km from the earth's surface. Kinetic energy of the satellite in this orbit is ___ ×1010J. (Mass of earth =6×1024kg, Radius of earth =6.4×106m, Gravitational constant =6.67×1011Nm2kg2)          [2025]



(3)

K=U2=GMm2r

 K=6.67×1011×6×1024×1032×6670×103=3×1010J