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Step Derivation/Formula Reasoning
1 g = \frac{G M}{R^2} The formula to calculate the gravitational field at the surface of a planet where G is the gravitational constant, M is the mass of the planet, and R is the radius of the planet.
2 g_{\text{Earth}} = \frac{G M_{\text{Earth}}}{R_{\text{Earth}}^2} Apply the gravitational field formula to Earth. Given that the gravitational field on Earth (standard gravity) is approximately 9.8 \, \text{m/s}^2 and R_{\text{Earth}} is approximately 6400 km or 6.4 \times 10^6 m.
3 M_{\text{new planet}} = \frac{1}{2} M_{\text{Earth}} The mass of the new planet is defined to be half that of Earth.
4 g_{\text{new planet}} = \frac{G (\frac{1}{2} M_{\text{Earth}})}{R_{\text{new planet}}^2} Substitute the mass of the new planet into the gravitational field formula.
5 20 = \frac{G (\frac{1}{2} M_{\text{Earth}})}{R_{\text{new planet}}^2} Given that g_{\text{new planet}} = 20 \, \text{N/kg}, we set this equal to the derived expression for gravitational field on the new planet.
6 g_{\text{Earth}} \cdot R_{\text{Earth}}^2 = 9.8 \cdot (6.4 \times 10^6)^2 Substitute the given values for Earth’s gravitational field and radius to calculate this product for further use.
7 \frac{1}{2} \cdot 9.8 \cdot (6.4 \times 10^6)^2 = 20 R_{\text{new planet}}^2 Equating the expressions from Earth and the new planet to find the radius of the new planet.
8 R_{\text{new planet}} = \sqrt{\frac{1}{2} \cdot \frac{9.8 \cdot (6.4 \times 10^6)^2}{20}} Solving for R_{\text{new planet}} using the simplified expression.
9 R_{\text{new planet}} \approx 3200 \, \text{km} The radius of the new planet is equal to half that of Earth because of the proportional division of mass and the square of the radius in calculating gravitational field strength.

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KinematicsForces
\Delta x = v_i t + \frac{1}{2} at^2F = ma
v = v_i + atF_g = \frac{G m_1m_2}{r^2}
a = \frac{\Delta v}{\Delta t}f = \mu N
R = \frac{v_i^2 \sin(2\theta)}{g}
Circular MotionEnergy
F_c = \frac{mv^2}{r}KE = \frac{1}{2} mv^2
a_c = \frac{v^2}{r}PE = mgh
KE_i + PE_i = KE_f + PE_f
MomentumTorque and Rotations
p = m v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum mr^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k x
T = 2\pi \sqrt{\frac{l}{g}}
T = 2\pi \sqrt{\frac{m}{k}}
ConstantDescription
gAcceleration due to gravity, typically 9.8 , \text{m/s}^2 on Earth’s surface
GUniversal Gravitational Constant, 6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2
\mu_k and \mu_sCoefficients of kinetic (\mu_k) and static (\mu_s) friction, dimensionless. Static friction (\mu_s) is usually greater than kinetic friction (\mu_k) as it resists the start of motion.
kSpring constant, in \text{N/m}
M_E = 5.972 \times 10^{24} , \text{kg} Mass of the Earth
M_M = 7.348 \times 10^{22} , \text{kg} Mass of the Moon
M_M = 1.989 \times 10^{30} , \text{kg} Mass of the Sun
VariableSI Unit
s (Displacement)\text{meters (m)}
v (Velocity)\text{meters per second (m/s)}
a (Acceleration)\text{meters per second squared (m/s}^2\text{)}
t (Time)\text{seconds (s)}
m (Mass)\text{kilograms (kg)}
VariableDerived SI Unit
F (Force)\text{newtons (N)}
E, PE, KE (Energy, Potential Energy, Kinetic Energy)\text{joules (J)}
P (Power)\text{watts (W)}
p (Momentum)\text{kilogram meters per second (kgm/s)}
\tau (Torque)\text{newton meters (Nm)}
I (Moment of Inertia)\text{kilogram meter squared (kgm}^2\text{)}
f (Frequency)\text{hertz (Hz)}

General Metric Conversion Chart

Example of using unit analysis: Convert 5 kilometers to millimeters.

1. Start with the given measurement: \text{5 km}

2. Use the conversion factors for kilometers to meters and meters to millimeters: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}

3. Perform the multiplication: \text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}

4. Simplify to get the final answer: \boxed{5 \times 10^6 \, \text{mm}}

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

10^{-12}

Nano-

n

10^{-9}

Micro-

µ

10^{-6}

Milli-

m

10^{-3}

Centi-

c

10^{-2}

Deci-

d

10^{-1}

(Base unit)

10^{0}

Deca- or Deka-

da

10^{1}

Hecto-

h

10^{2}

Kilo-

k

10^{3}

Mega-

M

10^{6}

Giga-

G

10^{9}

Tera-

T

10^{12}

1. Some answers may be slightly off by 1% depending on rounding, etc.
2. Answers will use different values of gravity. Some answers use 9.81 m/s2, and other 10 m/s2 for calculations.
3. Variables are sometimes written differently from class to class. For example, sometime initial velocity v_i is written as u ; sometimes \Delta x is written as s .
4. Bookmark questions that you can’t solve so you can come back to them later.
5. Always get help if you can’t figure out a problem. The sooner you can get it cleared up the better chances of you not getting it wrong on a test!

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