Out of the following,the only incorrect statement about satellites is

  • A
    $A$ satellite cannot move in a stable orbit in a plane passing through the earth's centre
  • B
    Geostationary satellites are launched in the equatorial plane
  • C
    We can use just one geostationary satellite for global communication around the globe
  • D
    The speed of a satellite increases with an increase in the radius of its orbit

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$A$ binary star system consists of two stars $A$ (mass $M_A = 2.2 M_S$) and $B$ (mass $M_B = 11 M_S$),where $M_S$ is the mass of the Sun. They are separated by a distance $d$ and rotate about their common centre of mass,which is stationary. What is the ratio of the angular momentum of star $A$ to the angular momentum of star $B$ about the centre of mass?

Suppose the acceleration due to gravity at the earth’s surface is $g \text{ m/s}^2$ and at the surface of the moon it is $g' \text{ m/s}^2$. An $M \text{ kg}$ passenger travels from the earth to the moon in a spaceship moving with a constant velocity. Neglecting all other celestial objects, which curve best represents the net gravitational force on the passenger as a function of time?

$A$ spacecraft travels in a straight line from the Earth to the Moon. How will its weight change during this journey?

In the following four periods:
$(i)$ Time of revolution of a satellite just above the earth's surface $({T_{st}})$
$(ii)$ Period of oscillation of mass inside the tunnel bored along the diameter of the earth $({T_{ma}})$
$(iii)$ Period of simple pendulum having a length equal to the earth's radius in a uniform field of $9.8 \; N/kg \; ({T_{sp}})$
$(iv)$ Period of an infinite length simple pendulum in the earth's real gravitational field $({T_{is}})$

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$A$ spherical body of radius $R$ consists of a fluid of constant density $\rho$ and is in equilibrium under its own gravity. If $P(r)$ is the pressure at a distance $r$ from the center $(r < R)$,then the correct option$(s)$ is(are):
$(A) P(r=0) = P_c$ (maximum pressure at center)
$(B) \frac{P(r=3R/4)}{P(r=2R/3)} = \frac{63}{80}$
$(C) \frac{P(r=3R/5)}{P(r=2R/5)} = \frac{16}{21}$
$(D) \frac{P(r=R/2)}{P(r=R/3)} = \frac{20}{27}$

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