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Gravitational force between masses and gravitational fields
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Every object in the universe attracts every other object with a gravitational force. Newton's Law of Universal Gravitation describes this attraction:
Calculate the gravitational force between Earth (mass = kg) and the Moon (mass = kg) when they are m apart.
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where:
๐ก Key Insight: Gravity acts between ALL objects, but we only notice it for very massive objects (like planets) because is extremely small!
Unlike electric force (which can attract or repel), gravitational force is always attractive.
By Newton's 3rd Law, the force Earth exerts on you equals the force you exert on Earth:
Both have magnitude
Force is inversely proportional to :
Gravity has infinite range (never becomes exactly zero), though it becomes negligibly small at large distances.
The gravitational field at a point is the gravitational force per unit mass:
where is the mass creating the field and is the distance from its center.
where:
At height above Earth's surface:
As altitude increases, decreases.
Example: At the altitude of the International Space Station (~400 km):
(About 89% of surface gravity - astronauts aren't "weightless" due to zero gravity, but due to free fall!)
Example: A 70 kg person
Use the distance from the center of the sphere:
This applies to any point outside the sphere (including at the surface).
Only the mass inside radius contributes to the force. The shell of mass outside has zero net effect!
For a uniform sphere:
So:
The force is proportional to inside a uniform sphere (increases linearly from zero at center).
When gravitational force provides the centripetal force, objects orbit!
Set gravitational force equal to required centripetal force:
Simplify (mass cancels):
The orbital speed is independent of the satellite's mass and depends only on:
Important: Closer orbits are faster! (Smaller means larger )
Using :
Kepler's 3rd Law:
The escape velocity is the minimum speed needed to escape a planet's gravitational pull:
For Earth: m/s (about 25,000 mph!)
Note: This is times the orbital velocity at the surface.
โ Wrong: โ Right:
โ Wrong: Using Earth's radius for a satellite 1000 km up โ Right: Use km
โ Wrong: "I'm weightless in space, so my mass is zero" โ Right: Mass is constant; weight is where varies with location
โ Wrong: "There's no gravity in space" โ Right: Gravity extends infinitely (though it weakens with distance). Astronauts feel "weightless" because they're in free fall, not because gravity is absent.
| Formula | Description |
|---|---|
| Gravitational force between masses | |
| Gravitational field strength | |
| Weight of object | |
Constants:
Given Information:
Find: Gravitational force
Solution:
Use Newton's Law of Universal Gravitation:
Substitute values:
Calculate the numerator:
Calculate the denominator:
Divide:
Answer: The gravitational force is approximately N (or 198 billion billion newtons).
Note: By Newton's 3rd Law, this is both the force Earth exerts on the Moon AND the force the Moon exerts on Earth!
The International Space Station orbits at an altitude of 400 km above Earth's surface. Calculate (a) the gravitational field strength at that altitude, and (b) the orbital speed of the ISS.
Given Information:
(a) Find gravitational field strength at altitude
Step 1: Find the distance from Earth's center
Step 2: Calculate gravitational field
Numerator:
Denominator:
(b) Find orbital speed
Step 3: Set gravitational force = centripetal force
The mass cancels:
Step 4: Calculate
Alternative for part (b):
Since , we have
Answers:
Note: Despite being nearly the same as on Earth's surface, astronauts feel "weightless" because both they and the ISS are in free fall toward Earth!
A satellite is in a circular orbit around Earth with a period of 24 hours (a geostationary orbit). Calculate: (a) the orbital radius, (b) the altitude above Earth's surface, and (c) the orbital speed.
Given Information:
(a) Find orbital radius
Step 1: Use the period formula
Step 2: Solve for
Square both sides:
Step 3: Calculate
Numerator:
Denominator:
Using calculator or estimation:
(b) Find altitude above Earth's surface
(c) Find orbital speed
Method 1: Using
Method 2: Using
Answers:
Application: Geostationary satellites at this altitude complete one orbit in exactly 24 hours, so they remain above the same point on Earth's equator - perfect for communication satellites!
| Orbital speed |
| Orbital period |
| Escape velocity |