 # The International Space Station has a mass of 4.2 x105 kg and orbits Earth at a distance of 4.0 x102 km above the surface. Earth has a radius of 6.37 x106 m, and mass of 5.97 x1024 kg. Calculate the following: The gravitational force between Earth and the ISS. The orbital speed of the ISS. The orbital period of the ISS (in minutes)

1. 3.65 x 106 N
2. 7,672 m/s
3. 92.41 minutes

## Written By Prof Phy AI

1. Gravitational Force Between Earth and the ISS

Step Formula Derivation Reasoning
1 F = G \frac{M_{\text{earth}} M_{\text{ISS}}}{r^2} Newton’s law of universal gravitation.
2 r = R_{\text{earth}} + h_{\text{ISS}} Distance r is Earth’s radius plus ISS’s altitude.

2. Orbital Speed of the ISS

Step Formula Derivation Reasoning
1 F_{\text{gravity}} = \frac{GM_{\text{earth}}M_{\text{ISS}}}{r^2} Gravitational force between Earth and ISS.
2 F_{\text{centripetal}} = \frac{M_{\text{ISS}}v^2}{r} Centripetal force required for circular orbit.
3 F_{\text{gravity}} = F_{\text{centripetal}} For a stable orbit, gravitational force equals centripetal force.
4 \frac{GM_{\text{earth}}M_{\text{ISS}}}{r^2} = \frac{M_{\text{ISS}}v^2}{r} Equating the two forces.
5 GM_{\text{earth}} = rv^2 Cancelling M_{\text{ISS}} and rearranging.
6 v = \sqrt{\frac{GM_{\text{earth}}}{r}} Solving for orbital velocity v. Note that is the total distance from the center of earth to ISS.

3. Orbital Period of the ISS

Step Formula Derivation Reasoning
1 T = \frac{\text{Circumference of Orbit}}{\text{Orbital Speed}} Orbital period T is the time to complete one orbit.
2 \text{Circumference} = 2\pi r Circumference formula for a circle.
3 T = \frac{2\pi r}{v} Substituting the circumference and orbital speed v.
4 T_{\text{minutes}} = \frac{T}{60} Converting period from seconds to minutes.

Let’s perform the calculations using the given values.

The calculations yield the following results:

1. Gravitational force between Earth and the ISS: \boxed{3,651,338, \text{N}}
2. Orbital speed of the ISS: \boxed{7,671.77, \text{m/s}}
3. Orbital period of the ISS: \boxed{92.41, \text{minutes}}

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1. 3.65 x 106 N
2. 7,672 m/s
3. 92.41 minutes ## Suggest an Edit

##### Nerd-Notes.com
KinematicsForces
\Delta x = v_i \cdot t + \frac{1}{2} a \cdot t^2F = m \cdot a
v = v_i + a \cdot tF_g = \frac{G \cdot m_1 \cdot m_2}{r^2}
a = \frac{\Delta v}{\Delta t}f = \mu \cdot N
R = \frac{v_i^2 \cdot \sin(2\theta)}{g}
Circular MotionEnergy
F_c = \frac{m \cdot v^2}{r}KE = \frac{1}{2} m \cdot v^2
a_c = \frac{v^2}{r}PE = m \cdot g \cdot h
KE_i + PE_i = KE_f + PE_f
MomentumTorque and Rotations
p = m \cdot v\tau = r \cdot F \cdot \sin(\theta)
J = \Delta pI = \sum m \cdot r^2
p_i = p_fL = I \cdot \omega
Simple Harmonic Motion
F = -k \cdot 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}
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 (kg·m/s)}
\tau (Torque)\text{newton meters (N·m)}
I (Moment of Inertia)\text{kilogram meter squared (kg·m}^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!