krit.club logo

Fields - Gravitational Fields

Grade 11IBPhysics

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Newton's Law of Universal Gravitation states that the gravitational force FF between two point masses MM and mm is directly proportional to the product of their masses and inversely proportional to the square of the separation distance rr between their centers.

•

The Gravitational Field Strength gg at a point is defined as the gravitational force per unit mass exerted on a small test mass placed at that point (g=Fmg = \frac{F}{m}).

•

Gravitational fields are represented by field lines; for a point mass or a uniform sphere, these lines are radial and point towards the center of the mass.

•

The gravitational constant GG has a value of approximately 6.67×10−11 N m2 kg−26.67 \times 10^{-11} \, \text{N} \, \text{m}^2 \, \text{kg}^{-2}.

•

Near the surface of a planet, the gravitational field is considered approximately uniform, meaning gg is constant in both magnitude and direction.

•

For an object in circular orbit, the gravitational force acts as the centripetal force, leading to the relationship GMmr2=mv2rG \frac{Mm}{r^2} = \frac{mv^2}{r}.

📐Formulae

F=GMmr2F = G \frac{Mm}{r^2}

g=Fmg = \frac{F}{m}

g=GMr2g = G \frac{M}{r^2}

v=GMrv = \sqrt{\frac{GM}{r}}

T2=4π2r3GMT^2 = \frac{4\pi^2 r^3}{GM}

💡Examples

Problem 1:

Calculate the gravitational field strength gg on the surface of Mars, given that the mass of Mars is 6.42×1023 kg6.42 \times 10^{23} \, \text{kg} and its radius is 3.39×106 m3.39 \times 10^6 \, \text{m}.

Solution:

g=GMR2g = G \frac{M}{R^2} g=(6.67×10−11)6.42×1023(3.39×106)2g = (6.67 \times 10^{-11}) \frac{6.42 \times 10^{23}}{(3.39 \times 10^6)^2} g≈3.73 m s−2g \approx 3.73 \, \text{m} \, \text{s}^{-2}

Explanation:

We use the formula for gravitational field strength at the surface of a spherical body. By substituting the universal gravitational constant, the mass of Mars, and the square of its radius, we find the acceleration due to gravity on its surface.

Problem 2:

A satellite orbits Earth at a height of 400 km400 \, \text{km} above the surface. If Earth's radius is 6.37×106 m6.37 \times 10^6 \, \text{m} and its mass is 5.97×1024 kg5.97 \times 10^{24} \, \text{kg}, determine the orbital speed vv of the satellite.

Solution:

r=REarth+h=6.37×106+0.40×106=6.77×106 mr = R_{Earth} + h = 6.37 \times 10^6 + 0.40 \times 10^6 = 6.77 \times 10^6 \, \text{m} v=GMrv = \sqrt{\frac{GM}{r}} v=(6.67×10−11)(5.97×1024)6.77×106v = \sqrt{\frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{6.77 \times 10^6}} v≈7670 m s−1v \approx 7670 \, \text{m} \, \text{s}^{-1}

Explanation:

The orbital radius rr must include both the Earth's radius and the altitude of the satellite. The orbital velocity is derived by equating the gravitational force to the centripetal force.

Gravitational Fields - Revision Notes & Key Formulas | IB Grade 11 Physics