Q.Derive an expression for critical velocity of a satellite.

Both orbital kinetic energy and total energy vary as $1/r$, but with opposite signs — so increasing the orbital radius makes KE smaller while making the (negative) total energy larger, i.e. closer to zero.

For a satellite of mass $m$ orbiting a planet of mass $M$ at radius $r$, gravity supplies the centripetal force:

$GMm/r^2 = mv^2/r \Rightarrow v^2 = GM/r$

Kinetic Energy:

$KE = (1/2)mv^2 = GMm/(2r)$

This is positive, and decreases as $r$ increases (larger orbit → slower orbital speed → smaller KE).

Potential Energy: $PE = -GMm/r$

Total Energy:

$TE = KE + PE = GMm/(2r) - GMm/r = -GMm/(2r)$

This is negative (the satellite is bound), and as $r$ increases, $-GMm/(2r)$ becomes less negative — i.e. it increases toward zero.

So as the orbital radius increases: KE decreases, T.E. increases.

Every body inside an orbiting artificial satellite experiences zero apparent weight — this is weightlessness.

An artificial satellite in orbit is in a state of continuous free fall around the Earth: gravity provides exactly the centripetal force needed to keep it in orbit, so both the satellite and any object inside it accelerate towards the Earth at the same rate. Since weight, as measured by a weighing scale, is really the normal reaction force the support exerts on a body, and here there is no relative acceleration between the body and its support (both fall together), the normal reaction — and hence the apparent weight — becomes zero. This is the state of weightlessness experienced by astronauts.

The orbital speed for a satellite in a near-surface circular orbit is v0 = sqrt(g*R), matching option (f).

For a satellite of mass m orbiting close to the Earth's surface (radius R) in a circular orbit, gravity provides the centripetal force:

mg = mv0^2/R

Solving for v0:

v0 = sqrt(g*R)

This is the minimum speed needed to maintain a stable circular orbit just above the Earth's surface (ignoring air resistance). Among the given options, only (f), sqrt(gR), matches this expression.

A geostationary satellite has a time period equal to Earth's own rotation period, ~24 hours.

A satellite is "geostationary" if it appears stationary relative to a point on Earth's surface — meaning it must complete one orbit in exactly the same time Earth takes to complete one rotation about its axis, moving in the same direction (west to east) in the equatorial plane.

Earth's rotation period is one day, so the geostationary satellite's orbital time period is also 24 hours (more precisely, one sidereal day, ≈ 23 h 56 min, but taken as 24 hours for standard NCERT-level purposes). This fixed period, via Kepler's third law, also fixes its orbital radius at about $4.2\times10^4$ km from Earth's centre.

Weightlessness in an orbiting satellite happens because the satellite (and everything in it) is in a state of continuous free fall -- gravity is not absent, but it is entirely used up as the centripetal force for the orbit.

For a satellite of mass m orbiting at radius r with orbital speed v, gravity supplies exactly the centripetal force needed:

GMm/r^2 = mv^2/r

This means the acceleration of the satellite (and everything inside it, including an astronaut) due to gravity equals the centripetal acceleration required for the orbit -- both the astronaut and the satellite accelerate towards Earth's centre at exactly the same rate. Since there is no relative acceleration between the astronaut and the satellite floor, the floor does not need to push up on the astronaut with any normal force. Weight, as we normally feel it, is actually this normal reaction force -- with it reduced to zero, the astronaut experiences apparent weightlessness, even though the actual gravitational force on them is not zero (it is only slightly less than at the Earth's surface, at typical orbital altitudes).

A synchronous satellite orbits the Earth once every 24 hours, matching the Earth's own rotation, so it appears stationary relative to a point on the surface (when placed in the equatorial plane).

Such a satellite's orbital radius is fixed by requiring its time period (from Kepler's third law, T = 2π√(r³/GM)) to equal Earth's sidereal rotation period of about 24 hours (≈23h 56m), which works out to an orbital radius of about 42,164 km from Earth's centre (~36,000 km altitude). This is the basis of geostationary communication satellites.

The satellite must complete one orbit in exactly the same time Earth takes to complete one rotation, so it stays fixed above the same point.

A geostationary satellite orbits Earth in the equatorial plane, in the same direction as Earth's rotation, and appears fixed relative to a point on Earth's surface. For this to be true, the satellite's orbital period must exactly match Earth's rotational period:

$T_{\text{satellite}} = T_{\text{Earth's rotation}} = 24\ \text{hours} \; (\approx 86400\ \text{s})$

This fixed period, combined with Kepler's third law, determines the unique orbital radius (~36,000 km altitude) at which a geostationary satellite must be placed.

A geostationary satellite has a 24-hour orbital period matched to Earth's rotation, so it always stays above the same point on the equator.

A geostationary satellite is an artificial satellite placed in a circular orbit in the plane of the Earth's equator, revolving in the same direction as the Earth's rotation (west to east), whose orbital period exactly equals the time the Earth takes to complete one rotation about its axis. Because the satellite and the point on the ground below it complete a revolution in the same time, the satellite appears to stay fixed at one position in the sky as seen from Earth. This makes it ideal for communication and weather-monitoring satellites, which must stay 'parked' over a fixed region. Its orbital radius works out to about $42{,}000\ km$ from Earth's centre.

A geostationary/communication satellite orbits with a period equal to Earth's rotation period, 24 hours, so it appears fixed over one point on Earth.

Communication satellites like the INSAT series are launched into a geostationary orbit — a circular orbit in the equatorial plane at an altitude of about 36,000 km, chosen so that the satellite's orbital angular velocity exactly matches Earth's rotational angular velocity. Because $T = 2\pi\sqrt{r^3/GM}$ increases with orbital radius, this specific radius is the one that makes $T$ equal to Earth's own rotation period. As a result the satellite appears stationary relative to a point on Earth's surface, which is essential for a fixed ground antenna to always point at it.

For a satellite skimming just above Earth's surface, gravity provides the centripetal force, giving $v_o = \sqrt{gR} \approx 7.92\text{ km/s}$.

For a satellite orbiting close to Earth's surface (orbital radius $\approx R$, Earth's radius), Newton's second law for circular motion gives:

$\frac{mv_o^2}{R} = \frac{GMm}{R^2} = mg$

$v_o = \sqrt{gR}$

Using $g = 9.8\text{ m/s}^2$ and $R = 6.4\times10^6\text{ m}$:

$v_o = \sqrt{9.8 \times 6.4\times10^6} \approx \sqrt{6.27\times10^7} \approx 7.92\times10^3\text{ m/s}$

A synchronous (geostationary) satellite orbits with the same 24-hour period as Earth's rotation, so it stays fixed relative to a point on the ground — that's what lets it continuously relay signals between two fixed locations.

For a satellite to be useful as a continuous relay between two points on Earth's surface, it must remain at a fixed position in the sky as seen from the ground — otherwise ground antennas would need to constantly track it and it would only be usable for short windows.

This 'geostationary' condition is achieved when the satellite's orbital period around the Earth exactly equals the Earth's own rotational period about its axis (≈ 23 h 56 min, essentially 24 hours), AND the orbit is equatorial and circular at the specific altitude (~36,000 km) where Kepler's third law gives this period. Then the satellite revolves at the same angular rate the Earth spins, so it appears stationary above a fixed point on the equator — able to relay TV/communication signals continuously between distant parts of the world within its view.