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Part (a) – Finding the orbital speed.

Step Derivation/Formula Reasoning
1 r = \frac{50}{2} = 25 \text{km} = 25000 \text{m} The radius r of the orbit is half the diameter. Convert it to meters.
2 T = 11 \times 86400 s Convert period T from days to seconds. There are 86400 seconds in a day.
3 v = \frac{2\pi r}{T} The orbital speed v is calculated by dividing the circumference of the orbit by the orbital period.
4 v = \frac{2\pi \times 25000}{950400} m/s Substitute the values of r and T into the formula. Convert r from km to m by multiplying by 1000.
5 v \approx 0.165 m/s Simplifying the expression gives the orbital speed v .
6 \boxed{v \approx 0.165 \text{m/s}} This is the final value for the satellite’s orbital speed.

Part (b) – Finding the comet’s mass.

Step Derivation/Formula Reasoning
1 v^2 = \frac{GM}{r} The inwards gravitational force is equal to the centripetal force of the orbiting comet. In terms of Newtons law this can be expressed as \frac{GMm}{r^2} = \frac{mv^2}{r} , where M is the mass of the comet, m   is the mass of the satellite, and G is the gravitational constant.
2 M = \frac{rv^2}{G} Rearrange the formula to solve for the mass M of the comet.
3 M = \frac{25000 \times (0.165)^2}{6.674 \times 10^{-11}} kg Substitute the values of r and v into the formula, remembering that r is already converted to meters.
4 M \approx 1.02 \times 10^{13} kg Calculating the value gives the mass of the comet.
5 \boxed{M \approx 1.02 \times 10^{13} \text{kg}} This is the final value for the mass of the comet.

Part (c) – Finding the landing speed.

Step Derivation/Formula Reasoning
1 h = 25\,\text{km}\, -\,1.8 \, \text{km} = 23.2 \, \text{km} = 23200 \, \text{meters} The distance from the satellite to the center of the comet is 25km. Since the comet has an average diameter of 3.6 km (or a radius of 1.8 km). The distance from the satellite to the surface of comet is 23.2 km. Convert this to meters.
2 t = 7 \times 3600 s Convert the fall time t from hours to seconds.
3 g = \frac{GM}{r^2} Calculate the acceleration g due to the comet’s gravity using the values from the previous two parts. Set mg = \frac{GMm}{r^2} and solve for g .
4 g = \frac{(6.67\times 10^{-11}) \times (1.02 \times 10^{13})}{25000^2} Substitute values into the equation for gravitational acceleration.
5 g = 1.09 \times 10^{-6} \, \text{m/s}^2 Final value for g ,the acceleration due to gravity of the comet.
6 v^2_{\text{final}} = v^2_{\text{initial}} + 2a\Delta x Now that we have v_{initial}, \, a, \, \Delta x we can use a kinematic formula to solve for v_f .
7 \boxed{v_{\text{final}} \approx .735\, \text{m/s}} Plug in all known values and solve for v_f .

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1. v \approx 0.165 \text{m/s}
2. M \approx 1.02 \times 10^{13} \text{kg}
3. v_{\text{final}} \approx .735\, \text{m/s}

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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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