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Physics: Space and Astrophysics - Universal Gravitation, Gravitational Fields, and Satellite Orbits

Grade 8IB

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

🔑Concepts

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Newton's Law of Universal Gravitation states that every point mass attracts every other point mass by a force acting along the line intersecting both points. This force FF is proportional to the product of the two masses m1m_1 and m2m_2, and inversely proportional to the square of the distance rr between them.

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The Gravitational Field Strength gg is defined as the gravitational force exerted per unit mass on a small object placed at that point in the field. Its unit is Newtons per kilogram (N/kgN/kg) or meters per second squared (m/s2m/s^2).

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Mass is the amount of matter in an object and remains constant regardless of location. Weight is the force of gravity acting on that mass and varies depending on the local gravitational field strength: W=m×gW = m \times g.

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An Inverse Square Law relationship exists for gravity: if the distance between two objects is doubled (2r2r), the gravitational force between them decreases to one-fourth (14F\frac{1}{4}F) of its original value.

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Satellites are kept in orbit by the gravitational pull of the planet they orbit. This force acts as a centripetal force, constantly changing the satellite's direction to keep it in a circular or elliptical path.

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For a stable circular orbit, the orbital speed vv must be precisely balanced with the altitude. If the speed is too high, the satellite will escape orbit; if too low, it will fall back to the planet.

📐Formulae

F=Gm1m2r2F = G \frac{m_1 m_2}{r^2}

W=mgW = m g

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

v=2πrTv = \frac{2 \pi r}{T}

💡Examples

Problem 1:

An astronaut has a mass of 75 kg75 \text{ kg}. Calculate their weight on Earth (g=9.8 N/kgg = 9.8 \text{ N/kg}) and on the Moon (g=1.6 N/kgg = 1.6 \text{ N/kg}).

Solution:

Weight on Earth: WEarth=75×9.8=735 NW_{Earth} = 75 \times 9.8 = 735 \text{ N} Weight on Moon: WMoon=75×1.6=120 NW_{Moon} = 75 \times 1.6 = 120 \text{ N}

Explanation:

Mass remains constant at 75 kg75 \text{ kg} in both locations, but weight changes because the gravitational field strength gg is different.

Problem 2:

If the gravitational force between two planets is 1000 N1000 \text{ N}, what will the force be if the distance between them is tripled?

Solution:

According to the inverse square law, F∝1r2F \propto \frac{1}{r^2}. If rr becomes 3r3r, the force becomes: Fnew=132×1000=19×1000≈111.11 NF_{new} = \frac{1}{3^2} \times 1000 = \frac{1}{9} \times 1000 \approx 111.11 \text{ N}

Explanation:

Since the distance is tripled, the denominator in the gravitation formula (r2r^2) increases by a factor of 32=93^2 = 9, making the force 99 times weaker.

Problem 3:

A satellite orbits Earth at a distance of 42000 km42000 \text{ km} from the center of the Earth. It takes 24 hours24 \text{ hours} to complete one orbit. Calculate its orbital speed in km/h\text{km/h}.

Solution:

v=2πrTv = \frac{2 \pi r}{T} v=2×3.14159×4200024v = \frac{2 \times 3.14159 \times 42000}{24} v=263893.7824≈10995.57 km/hv = \frac{263893.78}{24} \approx 10995.57 \text{ km/h}

Explanation:

The distance traveled in one orbit is the circumference of the circle (2πr2 \pi r). Speed is distance divided by the time period TT of the orbit.