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CBSE Class 9 Gravitation — practice questions

Explore 50 free Gravitation questions in CBSE Class 9 Science. Open each question or exercise to read its content, reference answers and available explanation.

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Gravitation questions and exercises with answers

Showing questions 1–30 of 50.

The universal gravitational constant $G$ has the SI value approximately:Read the answer and worked solution →For a satellite of mass $m$ orbiting Earth at radius $r$ (from the centre), the orbital velocity is:Read the answer and worked solution →The escape velocity from the Earth's surface is approximately ($R = 6.4 \times 10^6$ m, $g = 9.8$ m/s²):Read the answer and worked solution →Newton's law of gravitation states the force between two masses is:Read the answer and worked solution →At the surface of Earth, acceleration due to gravity g equals:Read the answer and worked solution →At the centre of the Earth, the value of g is:Read the answer and worked solution →Escape velocity from Earth's surface is approximately:Read the answer and worked solution →The first cosmic velocity (orbital velocity at Earth's surface) is approximately:Read the answer and worked solution →At a height h above Earth's surface where h ≪ R, g' equals:Read the answer and worked solution →At depth d below Earth's surface, g' equals:Read the answer and worked solution →Gravitational potential energy of a mass m at the surface of Earth (taking U at infinity as zero):Read the answer and worked solution →Time period T of a satellite in orbit of radius r around Earth is (Kepler's 3rd law):Read the answer and worked solution →Acceleration due to gravity at the EQUATOR compared to that at the POLES:Read the answer and worked solution →Gravitational field strength I (also called g) at distance r from a point mass M:Read the answer and worked solution →For a satellite to be GEOSTATIONARY (stays over the same point on Earth), its orbital period must equal:Read the answer and worked solution →Two masses 6 kg and 3 kg are 2 m apart. Gravitational force between them (G = 6.67 × 10⁻¹¹):Read the answer and worked solution →If Earth shrinks to half its radius keeping mass constant, the new escape velocity becomes:Read the answer and worked solution →Kepler's second law states that the line joining a planet to the sun sweeps equal areas in equal times. This is equivalent to conservation of:Read the answer and worked solution →A satellite is in circular orbit at radius 2R from Earth's centre. Its orbital velocity in terms of v_surface = √(gR):Read the answer and worked solution →Total energy of a satellite in circular orbit equals:Read the answer and worked solution →Two planets orbit the same star with semi-major axes 1 AU and 4 AU. Ratio of their orbital periods T1 : T2 is:Read the answer and worked solution →Spot the value of g (gravity) that is wrongRead the answer and worked solution →The gravitational pull of the Earth on an object is its:Read the answer and worked solution →The acceleration due to gravity is denoted by:Read the answer and worked solution →The value of g near the Earth's surface is about:Read the answer and worked solution →The unit of g is the same as that of:Read the answer and worked solution →Weight is measured with a:Read the answer and worked solution →The gravitational force on an object acts:Read the answer and worked solution →A book on a table is pushed up by the:Read the answer and worked solution →The gravitational force on a mass m is written as:Read the answer and worked solution →

Worked practice questions

The universal gravitational constant $G$ has the SI value approximately:

  1. $6.67 \times 10^{-11}\,\mathrm{N\,m^2/kg^2}$
  2. $9.81\,\mathrm{m/s^2}$
  3. $6.022 \times 10^{23}\,\mathrm{/mol}$
  4. $1.6 \times 10^{-19}\,\mathrm{C}$
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Correct answer: A. $6.67 \times 10^{-11}\,\mathrm{N\,m^2/kg^2}$

Universal gravitational constant $G = 6.67 \times 10^{-11}$ N·m²/kg² in SI units, the same everywhere in the universe. The units come from Newton's law $F = G m_1 m_2 / r^2$: $\dfrac{\text{kg} \cdot \text{kg}}{\text{m}^2} \cdot [G] = \text{N}$ so $[G] = \text{N} \cdot \text{m}^2 / \text{kg}^2$. Option B is the local acceleration due to gravity $g$ (not constant). Option C is Avogadro. Option D is the electronic charge.

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For a satellite of mass $m$ orbiting Earth at radius $r$ (from the centre), the orbital velocity is:

  1. $v = \sqrt{\dfrac{GM}{r}}$
  2. $v = \sqrt{2 GM r}$
  3. $v = \dfrac{GM}{r^2}$
  4. $v = G M$
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Correct answer: A. $v = \sqrt{\dfrac{GM}{r}}$

Set gravitational force equal to centripetal force on the satellite: $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$ The $m$ cancels. Solve for $v$: $v^2 = \dfrac{GM}{r}$, so $v = \sqrt{\dfrac{GM}{r}}$. Orbital velocity at Earth's surface is about $7.9$ km/s (about 28,000 km/h). Option B looks similar but has a 2 inside the square root — that's the **escape velocity** $v_e = \sqrt{2GM/r}$, which is $\sqrt{2}$ times the orbital speed at the same radius. A textbook trap.

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The escape velocity from the Earth's surface is approximately ($R = 6.4 \times 10^6$ m, $g = 9.8$ m/s²):

  1. $7.9$ km/s
  2. $11.2$ km/s
  3. $20$ km/s
  4. $1000$ m/s
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Correct answer: B. $11.2$ km/s

Escape velocity from Earth's surface: $v_e = \sqrt{2gR}$ (using the surface relation $GM/R^2 = g$). $v_e = \sqrt{2 \times 9.8 \times 6.4 \times 10^6} = \sqrt{1.25 \times 10^8} \approx 1.12 \times 10^4$ m/s $= 11.2$ km/s. Option A ($7.9$ km/s) is the *orbital* velocity at Earth's surface, smaller by a factor of $\sqrt{2}$. That trap is identical to the one in the orbital-velocity question.

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