Gravitation
Maharashtra Board · Class 11 · Physics
Practice quiz for Gravitation — Maharashtra Board Class 11 Physics. MCQs and questions with answers to test your preparation.
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Quick Quiz: Gravitation
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What is the value of universal gravitational constant G in SI units?
According to Kepler's first law, planets move in which type of orbit around the Sun?
A satellite is orbiting Earth at a height of 400 km above the surface. If Earth's radius is 6400 km and g = 10 m/s², what is the orbital velocity of the satellite?
What is the escape velocity from Earth's surface? (Given: g = 10 m/s², R = 6.4 × 10⁶ m)
Sample Questions
Which of the following factors affect the acceleration due to gravity on Earth's surface? (Select all correct answers)
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Mass of the Earth, Radius of the Earth, Rotation of the Earth, Latitude, Shape of the Earth
Acceleration due to gravity g = GM/R² depends on Earth's mass M and radius R. Earth's rotation causes centrifugal effect reducing apparent g. Latitude affects g due to rotation and Earth's oblate shape. The mass of the object cancels out in the acceleration formula, so it doesn't affect g.
At what height above Earth's surface does the acceleration due to gravity become one-fourth of its surface value?
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R (one Earth radius)
Step-by-step solution: At height h: gₕ = GM/(R+h)² At surface: g = GM/R² Given: gₕ = g/4 Therefore: GM/(R+h)² = (1/4) × GM/R² (R+h)² = 4R² R+h = 2R h = R So at height equal to Earth's radius, gravity becomes one-fourth.
Which statement about geostationary satellites is correct?
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They orbit Earth in 24 hours
Geostationary satellites have an orbital period of 24 hours, which matches Earth's rotational period. This makes them appear stationary relative to Earth's surface. They orbit in the equatorial plane at a specific altitude of about 36,000 km above Earth's surface.
The gravitational force between two masses is 100 N when they are 10 m apart. What will be the force when the distance is doubled?
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25 N
Step-by-step solution: Gravitational force F = Gm₁m₂/r² Initially: F₁ = 100 N at r₁ = 10 m Finally: r₂ = 2r₁ = 20 m F₂ = Gm₁m₂/(2r₁)² = Gm₁m₂/(4r₁²) = F₁/4 = 100/4 = 25 N When distance doubles, force becomes one-fourth due to inverse square law.
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