Is it correct to say: Using h alone instead of R + h in v_c = √(GM/(R + h))?

The orbit radius is measured from the Earth's centre: r = R + h. For h = 400 km, r = 6800 km, not 400 km.

Is it correct to say: Substituting R and h in km while G is in SI units?

Convert to metres first: 6400 km = 6.4 × 10⁶ m. The answer then comes out in m/s.

Is it correct to say: Thinking a heavier satellite needs a larger orbital speed?

The satellite's mass m cancels. v_c depends only on M and R + h.

Is it correct to say: Confusing escape velocity with critical velocity?

Critical velocity keeps a satellite in orbit (≈ 7.9 km/s near the surface). Escape velocity takes it out of the Earth's gravity altogether (≈ 11.2 km/s). v_esc = √2 × v_c.

Is it correct to say: Saying a geosynchronous satellite is at rest?

It moves at about 3 km/s, but its period equals the Earth's rotation period (about 24 h), so when it orbits above (parallel to) the equator it appears stationary from the ground.

Is it correct to say: Leaving the period in seconds when hours or minutes are asked?

Divide by 60 for minutes and by 3600 for hours. 5570 s ≈ 93 min.

📄 Formula sheet (PDF)

Space Missions

Artificial satellites, critical (orbital) velocity, period of a satellite, classification of satellite orbits, escape velocity and missions away from the Earth — Maharashtra SSC Science-1 Ch 10

📐 10 formulas✏️ 3 examples🎯 6 practice🏫 Maharashtra State Board📚 Class 10✓ 2025–26 syllabus

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Board Exam Tips

→The derivation of critical velocity (equate gravitational force to the centripetal force) is short and worth learning step by step.
→Always use r = R + h, measured from the Earth's CENTRE. Convert km to m before substituting.
→Learn the orbit table: height range, an example use and why satellites in that orbit are used for it.
→Remember the standard values used in the textbook: G = 6.67 × 10⁻¹¹ N m²/kg², M = 6 × 10²⁴ kg, R = 6400 km.
→Ratio questions (what happens to v or T if the orbit radius changes) are fastest with v ∝ 1/√r and T² ∝ r³ — no need to substitute G and M.

📐 Formulas(10)

Condition for a Circular Orbit

\frac{m v _{c}^{2}}{R + h} = \frac{GM m}{( R + h ) ^{2}}

Symbol	Meaning
$m$	Mass of the satellite (kg)
$M$	Mass of the Earth = 6 × 10²⁴ kg
$R$	Radius of the Earth = 6.4 × 10⁶ m
$h$	Height of the satellite above the Earth's surface (m)

Critical Velocity of a Satellite★ Board fav

v_{c} = \frac{GM}{R + h}

Symbol	Meaning
$v_{c}$	Critical (orbital) velocity (m/s)
$G$	Universal gravitational constant = 6.67 × 10⁻¹¹ N m²/kg²
$R + h$	Radius of the orbit, from the Earth's centre (m)

Critical Velocity Just Above the Surface

v_{c} = \frac{GM}{R} = g R \approx 7.9 km/s

Period of Revolution of a Satellite★ Board fav

T = \frac{2 π ( R + h )}{v _{c}}

Symbol	Meaning
$T$	Period of revolution (s)

Period in Terms of Orbit Radius

T = 2 π \frac{( R + h ) ^{3}}{GM}

Kepler's Third Law for Satellites

T^{2} \propto (R + h)^{3}

Dependence of Speed on Orbit Radius

v_{c} \propto \frac{1}{R + h}

Escape Velocity★ Board fav

v_{esc} = \frac{2 GM}{R} = 2 g R

Symbol	Meaning
$v_{esc}$	Escape velocity (m/s)
$R$	Radius of the planet (m)

Escape Velocity and Critical Velocity

v_{esc} = 2 v_{c} (near the surface)

Classification of Satellite Orbits

LEO: 180 to 2000 km, MEO: 2000 to 35780 km, HEO: h \geq 35780 km

✏️ Solved Examples

1Solved Exampleeasy3 steps

A planet has mass 6.4 × 10²³ kg and radius 3.4 × 10⁶ m. Find the escape velocity from its surface. (G = 6.67 × 10⁻¹¹ N m²/kg²)

Escape velocity formula.

v_{esc} = \frac{2 GM}{R} = \frac{2 \times 6.67 \times 1 0 ^{- 11} \times 6.4 \times 1 0 ^{23}}{3.4 \times 1 0 ^{6}}

2Solved Exampleboard4 steps

A satellite revolves in a circular orbit 400 km above the Earth's surface. Find its critical velocity and period of revolution. (G = 6.67 × 10⁻¹¹ N m²/kg², M = 6 × 10²⁴ kg, R = 6400 km)

Radius of the orbit from the Earth's centre.

R + h = 6400 + 400 = 6800 km = 6.8 \times 1 0^{6} m

3Solved ExampleHOTS4 steps

Satellite A orbits the Earth at a height equal to the Earth's radius R. Satellite B has a period 8 times that of A. Find the height of B above the surface and compare the speeds of A and B.

Orbit radius of A, from the Earth's centre.

r_{A} = R + R = 2 R

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Using h alone instead of R + h in v_c = √(GM/(R + h))
✓The orbit radius is measured from the Earth's centre: r = R + h. For h = 400 km, r = 6800 km, not 400 km.
2
Substituting R and h in km while G is in SI units
✓Convert to metres first: 6400 km = 6.4 × 10⁶ m. The answer then comes out in m/s.
3
Thinking a heavier satellite needs a larger orbital speed
✓The satellite's mass m cancels. v_c depends only on M and R + h.
4
Confusing escape velocity with critical velocity
✓Critical velocity keeps a satellite in orbit (≈ 7.9 km/s near the surface). Escape velocity takes it out of the Earth's gravity altogether (≈ 11.2 km/s). v_esc = √2 × v_c.
5
Saying a geosynchronous satellite is at rest
✓It moves at about 3 km/s, but its period equals the Earth's rotation period (about 24 h), so when it orbits above (parallel to) the equator it appears stationary from the ground.
6
Leaving the period in seconds when hours or minutes are asked
✓Divide by 60 for minutes and by 3600 for hours. 5570 s ≈ 93 min.

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
Find the critical velocity of a satellite moving just above the Earth's surface. (g = 9.8 m/s², R = 6.4 × 10⁶ m) Compare it with the escape velocity.
Q2
Find the critical velocity and period of a satellite at a height of 1600 km. (G = 6.67 × 10⁻¹¹ N m²/kg², M = 6 × 10²⁴ kg, R = 6400 km)
Q3
Classify these orbits as low, medium or high Earth orbit: (i) 500 km (ii) 20,200 km (iii) 35,780 km.
Q4
If the radius of a satellite's orbit is made 9 times larger, how do its speed and period change?
Q5
A planet has 4 times the mass and 2 times the radius of the Earth. Find the escape velocity from it. (v_esc for Earth = 11.2 km/s)
Q6
Using M = 6 × 10²⁴ kg, R = 6400 km and h = 35780 km, find the critical velocity and period of a satellite in high Earth orbit.

📝 Notes

Space Missions

Artificial satellites are man-made objects placed in orbit around the Earth using launch vehicles. Whether a satellite stays in orbit, falls back, or leaves the Earth altogether depends only on its speed and its distance from the Earth's centre.

Why a satellite does not fall

A satellite is falling towards the Earth all the time, but it also moves sideways fast enough that the Earth's surface curves away beneath it. In a circular orbit, gravity supplies exactly the centripetal force: $\frac{mv_c^2}{R+h} = \frac{GMm}{(R+h)^2}$ , which gives the critical velocity $v_c = \sqrt{GM/(R+h)}$ .

Less than $v_c$ ⇒ it cannot stay in that orbit and falls back towards the Earth.
Exactly $v_c$ ⇒ circular orbit.
At or above $v_{esc} = \sqrt{2GM/R}$ from the surface ⇒ it leaves the Earth's gravity.

The orbit table

Orbit	Height above surface	Typical use
Low Earth orbit	180 – 2000 km	Scientific experiments, atmospheric studies; International Space Station, Hubble telescope
Medium Earth orbit	2000 – 35780 km	Polar orbits (study of polar regions); GPS satellites at about 20,200 km
High Earth orbit	35780 km and above	Geosynchronous satellites: meteorology, telephone, TV and radio signals

At about 35780 km the period becomes 24 hours, equal to the Earth's rotation, so a satellite above the equator appears fixed in the sky.

Missions away from the Earth

To leave the Earth's gravity altogether — for example, on a mission to another planet — a spacecraft must reach the Earth's escape velocity (about 11.2 km/s from the surface). Know the purpose of the Indian missions described in the textbook, such as Chandrayaan-1 to the Moon and the Mars Orbiter Mission (Mangalyaan).

Space debris

Non-working satellites, spent rocket stages and fragments keep orbiting the Earth. This space debris can collide with working satellites, so its tracking and removal is an important concern.