IB MYP 4-5 Physics- Weight and gravitational force - Study Notes - New Syllabus
IB MYP 4-5 Physics-Weight and gravitational force – Study Notes
Key Concepts
- Weight and gravitational force
Weight and Gravitational Force
Concept of Weight:
Weight is the force exerted on a body due to the gravitational attraction of a massive object (e.g., Earth). It depends on both the mass of the object and the gravitational field strength at the location of the object.
$\mathrm{W = m g}$
- \( W \) = Weight (N)
- \( m \) = Mass (kg)
- \( g \) = Gravitational field strength (\( \mathrm{N\,kg^{-1}} \))
Important Notes:
- On Earth, \( g \approx 9.81 \ \mathrm{m/s^2} \) but can vary slightly with altitude and location.
- Weight changes with \( g \), while mass remains constant regardless of location.
- On the Moon, \( g \) is about \( 1.62 \ \mathrm{m/s^2} \), so weight is much less for the same mass.
Gravitational Force:
Any two masses attract each other with a force given by Newton’s Law of Gravitation:
$F = G \dfrac{m_1 m_2}{r^2}$
- \( F \) = gravitational force (N)
- \( G \) = universal gravitational constant = \( 6.674 \times 10^{-11} \ \mathrm{N \cdot m^2 / kg^2} \)
- \( m_1, m_2 \) = masses of the two objects (kg)
- \( r \) = distance between their centers (m)
Key Points:
- Near Earth’s surface, this force is simplified to \( W = mg \).
- At large distances, the inverse-square law applies directly.
Example:
Find the weight of a 75 kg astronaut on Earth and on the Moon.
▶️ Answer/Explanation
On Earth:
W_E = m g_E = 75 \times 9.81 = 735.75 \ \mathrm{N}
On Moon:
W_M = m g_M = 75 \times 1.62 = 121.5 \ \mathrm{N}
Final Answer:
Earth: \(\boxed{735.75 \ \mathrm{N}}\), Moon: \(\boxed{121.5 \ \mathrm{N}}\)
Example:
Two objects of masses \( m_1 = 10 \ \mathrm{kg} \) and \( m_2 = 5 \ \mathrm{kg} \) are placed 2 m apart in space. Find the gravitational force between them.
▶️ Answer/Explanation
Using \( F = G \dfrac{m_1 m_2}{r^2} \):
F = (6.674 \times 10^{-11}) \dfrac{(10)(5)}{(2)^2}
F = (6.674 \times 10^{-11}) \dfrac{50}{4}
F = (6.674 \times 10^{-11}) \times 12.5
F = 8.3425 \times 10^{-10} \ \mathrm{N}
Final Answer: \(\boxed{8.34 \times 10^{-10} \ \mathrm{N}}\)
Orbital Speed and Escape Velocity
Orbital Speed
The orbital speed is the speed at which an object must move to stay in a stable orbit around a planet or star.It occurs when the gravitational force provides the exact centripetal force needed to keep the object moving in a circular path.
Formula:
\(v_\text{orb} = \sqrt{\dfrac{GM}{r}}\)
- \(v_\text{orb}\) = orbital speed
- \(G\) = gravitational constant (\(6.674 \times 10^{-11} \, \mathrm{N\,m^2/kg^2}\))
- \(M\) = mass of the central body (planet or star)
- \(r\) = distance from the center of the body to the orbiting object
Objects at orbital speed follow a curved path around the planet without falling into it or escaping into space.
Example:
The International Space Station orbits Earth at approximately 7.7 km/s at a height of ~400 km.
Escape Velocity
Escape velocity is the minimum speed needed for an object to break free from the gravitational pull of a planet or star without further propulsion. At escape velocity, kinetic energy equals the gravitational potential energy at that distance from the center of the body.
Formula:
\(v_\text{esc} = \sqrt{\dfrac{2GM}{r}}\)
- \(v_\text{esc}\) = escape velocity
- \(G\) = gravitational constant
- \(M\) = mass of the central body
- \(r\) = distance from the center of the body to the object
Escape velocity is always higher than orbital speed by a factor of \(\sqrt{2}\) for the same radius.
Example: For Earth, escape velocity at the surface is approximately 11.2 km/s.
Example:
Calculate the orbital speed of a satellite orbiting Earth at a height of 400 km. (Mass of Earth \(M = 5.97 \times 10^{24} \, \mathrm{kg}\), radius of Earth \(R = 6371 \, \mathrm{km}\))
▶️ Answer/Explanation
Step 1 – Find distance from Earth’s center: \(r = R + h = 6371 + 400 = 6771 \, \mathrm{km} = 6.771 \times 10^6 \, \mathrm{m}\)
Step 2 – Apply orbital speed formula: \(v_\text{orb} = \sqrt{\dfrac{GM}{r}}\)
\(v_\text{orb} = \sqrt{\dfrac{6.674\times10^{-11} \times 5.97\times10^{24}}{6.771\times10^6}} \approx 7.67 \, \mathrm{km/s}\)
Answer: \(\boxed{v_\text{orb} \approx 7.7 \, \mathrm{km/s}}\)
Example :
Calculate the escape velocity from Earth’s surface. (Use the same values as above Example)
▶️ Answer/Explanation
Step 1 – Apply escape velocity formula: \(v_\text{esc} = \sqrt{\dfrac{2GM}{r}}\)
\(v_\text{esc} = \sqrt{\dfrac{2 \times 6.674\times10^{-11} \times 5.97\times10^{24}}{6.371\times10^6}} \approx 11.2 \, \mathrm{km/s}\)
Answer: \(\boxed{v_\text{esc} \approx 11.2 \, \mathrm{km/s}}\)