Kit Library / Science / Physics
⚡ Topic Learning KitShared by a Veda learner · Generated with Veda AI
Veda Bites are swipeable micro-lessons — each one teaches exactly one idea. The bites below come from this kit — swipe through them in the app.
Gravity is universal — not just Earth's.
Newton's law of gravitation states that every particle of matter attracts every other particle with a force directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
↳ Gravitation is a universal, always-attractive, action-at-a-distance force between any two masses.
One number rules all gravity.
↳ G is a universal constant — its tiny value explains why gravity feels weak between everyday objects.
G and g are NOT the same.
↳ G is a fixed universal constant; g is a local value that changes with position.
Why g equals 9.8 m/s².
Equating gravitational force to weight $mg$ gives the surface value of g.
↳ g depends only on the mass and radius of the planet — not on the mass of the falling object.
Feather vs coin — same g, different drag.
↳ g is independent of the falling object's mass — heavier objects don't fall faster in vacuum.
g shrinks as you climb or dig.
↳ g decreases with height (≈ 1 − 2h/R) and decreases linearly with depth (1 − d/R), becoming zero at Earth's centre.
Planets follow three elegant rules.
↳ Kepler's laws are empirical; Newton's law of gravitation explains all three.
The minus sign matters here.
For two masses separated by distance r, gravitational potential energy is taken as zero at infinity.
↳ U is always negative for a bound system — energy must be supplied to separate masses to infinity.
11.2 km/s and you're gone.
↳ Escape velocity is the minimum speed needed to leave a planet's gravity permanently — ~11.2 km/s for Earth.
The speed that keeps you circling.
For a satellite in a circular orbit of radius r, gravitational force provides the centripetal force.
↳ Orbital velocity decreases with orbital radius; near Earth it's about 7.9 km/s.
They're falling — just missing Earth.
↳ Weightlessness in orbit = free fall, not zero gravity.
Gravity runs the solar system and your GPS.
↳ Gravitation is not abstract — it powers tides, satellites, and space exploration.
Memorise these five — they cover most questions.
↳ These five formulas solve the majority of numericals in this chapter.
Escape is √2 times orbital.
Since $v_e = \sqrt{2GM/R}$ and $v_o = \sqrt{GM/R}$, we get $v_e = \sqrt{2}\, v_o$ — escape velocity is always √2 (≈1.41) times the near-surface orbital velocity.
↳ v_e = √2 · v_o — a quick ratio to remember for numericals.
Yet it shapes the cosmos.
Gravitation is by far the weakest of the four fundamental forces — but because it is always attractive and has infinite range, it dominates at cosmic scales, shaping galaxies, stars, and planets.
↳ Weak locally, dominant cosmically — gravity's range and universality make it the architect of the universe.
Gravitation is the universal attractive force between any two masses, governed by Newton's law of gravitation. This chapter covers the universal law, acceleration due to gravity, variation of g with height and depth, Kep…
Full explanation, worked examples & memory tricks in the app
All 17 questions, laddered from warm-up to topper-level, each with its answer and the reasoning behind it.
Q1. Which statement best describes the nature of gravitational force?
BeginnerA · It is always attractive and acts along the line joining the masses.
Gravitational force is always attractive and acts along the line joining the two masses; it never repels and needs no contact.
Q2. The value of the universal gravitational constant $G$ is:
BeginnerC · $6.67 \times 10^{-11}\ \text{N m}^2\text{kg}^{-2}$
$G = 6.67 \times 10^{-11}\ \text{N m}^2\text{kg}^{-2}$ is the universal gravitational constant. $9.8\ \text{m s}^{-2}$ is $g$ on Earth and $11.2\ \text{km/s}$ is escape velocity.
Q3. Two masses are separated by a distance $r$. If the distance is tripled, the gravitational force becomes:
BeginnerB · One-ninth of the original force
Since $F \propto 1/r^2$, tripling $r$ makes the force $1/3^2 = 1/9$ of its original value.
Q4. A student says, "My weight is 60 kg." What is wrong with this statement?
BeginnerD · Weight is a force measured in newtons; 60 kg is the mass.
Mass is the quantity of matter in kg and is constant; weight is the gravitational force $W = mg$ measured in newtons and changes with $g$.
Q5. The universal gravitational constant $G = 6.67 \times 10^{-11}\ \text{N m}^2\text{kg}^{-2}$ is the same everywhere in the universe.
BeginnerA · True
The universal gravitational constant $G = 6.67 \times 10^{-11}\ \text{N m}^2\text{kg}^{-2}$ is the same everywhere in the universe.
Q6. A satellite orbits Earth at radius $r$ from Earth's centre. Which expression gives its orbital velocity?
IntermediateD · $v = \sqrt{\frac{GM}{r}}$
Equating gravitational force to centripetal force, $\frac{GMm}{r^2} = \frac{mv^2}{r}$, gives $v = \sqrt{GM/r}$. The factor of 2 appears only in escape velocity.
Q7. Why is $G$ called the universal gravitational constant?
IntermediateB · Its value is the same everywhere in the universe.
$G$ is a fundamental constant with the same value everywhere. In contrast, $g = GM/R^2$ is local and varies with the planet's mass and radius.
Q8. An astronaut in an orbiting space station feels weightless. Which explanation is correct?
IntermediateB · The astronaut and station are in free fall together, so the normal force is zero.
Apparent weightlessness is free fall, not absence of gravity. Both astronaut and station accelerate toward Earth at the same rate, so no normal force acts.
Q9. For a body on Earth's surface, which equation correctly links $g$ to $G$, Earth's mass $M$, and Earth's radius $R$?
IntermediateB · $g = \frac{GM}{R^2}$
Equating $mg$ with $GMm/R^2$ and cancelling $m$ gives $g = GM/R^2$.
Q10. Which of the following is the correct expression for gravitational potential energy at distance $r$ from a mass $M$?
IntermediateD · $U = -\frac{GMm}{r}$
The exact form is $U = -GMm/r$, with the negative sign indicating a bound system. $U = mgh$ is only a small-height approximation.
Q11. Earth's escape velocity is about:
IntermediateB · $11.2\ \text{km/s}$
Using $v_e = \sqrt{2GM/R}$ for Earth gives about $11.2\ \text{km/s}$. $2.4\ \text{km/s}$ is the Moon's escape velocity and $3 \times 10^8\ \text{m/s}$ is the speed of light.
Q12. Kepler's third law states that:
IntermediateB · $T^2 \propto a^3$
Kepler's third law is $T^2 \propto a^3$, which follows from equating gravitational and centripetal forces for an orbit.
Q13. Which statement about the four fundamental forces is correct?
AdvancedA · Gravity is the weakest but dominates on astronomical scales because it is always attractive and has infinite range.
Gravity is the weakest fundamental force, but because it is always attractive and has infinite range, it accumulates over huge masses and governs astronomical structures.
Q14. A planet has twice Earth's mass and twice Earth's radius. Its surface gravity compared to Earth's is:
AdvancedA · Half of Earth's $g$
$g = GM/R^2$. Doubling both $M$ and $R$ gives $g' = G(2M)/(2R)^2 = (2/4)\,GM/R^2 = g/2$.
Q15. Why does the Moon not fall into Earth despite gravity pulling it?
AdvancedB · Its tangential velocity keeps it in orbit; gravity acts as the centripetal force.
The Moon continuously falls toward Earth, but its tangential velocity carries it around; gravity provides the centripetal force $GMm/r^2 = mv^2/r$.
Q16. Assertion (A): A black hole traps light. Reason (R): Its escape velocity exceeds the speed of light. Choose the correct option.
AdvancedB · Both A and R are true, and R correctly explains A.
A black hole's escape velocity $v_e = \sqrt{2GM/R}$ exceeds $c$, so even light cannot escape — R is the correct explanation of A.
Q17. A satellite is in a circular orbit around Earth at a height of 500 km above the surface. Calculate its orbital velocity. Given that the radius of Earth is approximately 6400 km and the universal gravitational constant is \( G = 6.67 imes 10^{-11} \ ext{N m}^2 ext{kg}^{-2} \) and Earth's mass \( M \approx 5.97 imes 10^{24} \ ext{kg} \).
AdvancedA · 7.91 km/s
To find the orbital velocity \( v \) of the satellite, we use the formula \( v = \sqrt{ rac{GM}{r}} \). The radius \( r \) is the sum of Earth's radius and the height of the satellite: \( r = 6400 \ ext{km} + 500 \ ext{km} = 6900 \ ext{km} = 6.9 imes 10^6 \ ext{m} \). Pluggi…
20 flashcards in this kit — the app reviews them with spaced repetition so the right card returns on the right day.
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Word Match
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True False
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Memory Match
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Flashcard Battle
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Speed Quiz
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Guess Term
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Sequence Builder
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Categorization
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Revision Battle
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