Login

AP Physics

Unit 3 - Circular Motion

13450 views

Advanced

Mathematical

GQ

You're a Pro Member

Supercharge UBQ

Go Pro

0 attempts

0% avg

UBQ Credits

Verfied Answer
Verfied Explanation 0 likes
Written by Phy
0

Escape speed is the minimum speed needed for an object to break free from the gravitational pull of a planet without further propulsion. The formula for escape speed ([katex] v_{\text{escape}} [/katex]) is derived from setting the kinetic energy equal to the gravitational potential energy. However, the following is how to do it without using energy:

Step Formula Derivation Reasoning
1 [katex] F = \frac{GMm}{r^2} [/katex] Gravitational force formula, where [katex] G [/katex] is the gravitational constant, [katex] M [/katex] is the mass of the planet, [katex] m [/katex] is the mass of the object, and [katex] r [/katex] is the distance from the center of the planet.
2 [katex] a = \frac{F}{m} = \frac{GM}{r^2} [/katex] Acceleration due to gravity, derived from the gravitational force (Newton’s second law).
3 [katex] v^2 = u^2 + 2as [/katex] Kinematic equation for velocity, where [katex] v [/katex] is the final velocity, [katex] u [/katex] is the initial velocity, [katex] a [/katex] is the acceleration, and [katex] s [/katex] is the displacement.
4 [katex] v_{\text{escape}}^2 = 2 \times \frac{GM}{r} [/katex] Setting [katex] u = 0 [/katex] (initial speed), [katex] v = 0 [/katex] (final speed at infinity), and [katex] s = r [/katex] (escape from the surface), and solving for [katex] v_{\text{escape}} [/katex].
5 [katex] v_{\text{escape}} = \sqrt{2 \times \frac{GM}{r}} [/katex] Calculating escape velocity.

Given:

  • Mass of the planet ([katex] M [/katex]): [katex] 6.89 \times 10^{25} , \text{kg} [/katex]
  • Radius of the planet ([katex] r [/katex]): [katex] 6.2 \times 10^{6} , \text{m} [/katex]

We can now compute the escape velocity.

Step Formula Derivation Reasoning
5 [katex] v_{\text{escape}} \approx 38515 , \text{m/s} [/katex] Calculated escape velocity.

The escape speed from a planet with a mass of [katex] 6.89 \times 10^{25} , \text{kg} [/katex] and a radius of [katex] 6.2 \times 10^{6} , \text{m} [/katex] is approximately [katex] \boxed{38515 , \text{m/s}} [/katex], calculated without using energy concepts.

Need Help? Ask Phy To Explain

Just ask: "Help me solve this problem."

Just Drag and Drop!
Quick Actions ?
×

Topics in this question

Join 1-to-1 Elite Tutoring

  • Statistics

See how Others Did on this question | Coming Soon

Intermediate
Conceptual
MCQ

A girl throws a stone from a bridge. Consider the following ways she might throw the stone. The speed of the stone as it leaves her hand is the same in each case.

Case A: Thrown straight up.

Case B: Thrown straight down.

Case C: Thrown out at an angle of 45° above horizontal.

Case D: Thrown straight out horizontally.

In which case will the speed of the stone be greatest when it hits the water below if there is no significant air resistance, assuming equal initial speeds?

  • Energy
Solve
Advanced
Proportional Analysis
MCQ

A rocket of mass m is launched with kinetic energy K0, from the surface of the Earth. How much less kinetic energy does the rocket have at an altitude of two Earth radii in terms of the gravitational constant, G; the mass of the Earth, mE ; the radius of the Earth, RE ; and the mass of the rocket?

  • Energy
Solve
Intermediate
Conceptual
MCQ

A rubber ball bounces off of a wall with an initial speed v and reverses its direction so its speed is v right after the bounce. As a result of this bounce, which of the following quantities of the ball are conserved?

  • Energy, Momentum
Solve
Intermediate
Mathematical
MCQ

A pendulum consists of a ball of mass [katex] m [/katex] suspended at the end of a massless cord of length [katex] L [/katex]. The pendulum is drawn aside through an angle of 60° with the vertical and released. At the low point of its swing, the speed of the pendulum ball is

  • Centripetal Acceleration, Circular Motion, Energy, Pendulums, Simple Harmonic Motion
Solve
Advanced
Mathematical
GQ

A [katex] 2.2 \times 10^{21} \, \text{kg}[/katex] moon orbits a distant planet in a circular orbit of radius [katex] 1.5 \times 10^8 \, \text{m}[/katex]. It experiences a [katex] 1.1 \times 10^{19} \, \text{N}[/katex] gravitational pull from the planet. What is the moon’s orbital period in earth days?

  • Circular Motion, Gravitation
Solve
Discussion Threads
Login to Discuss

Leave a Reply

v = 38,515 m/s

Nerd Notes

Discover the world's best Physics resources

Continue with

By continuing you (1) agree to our Terms of Sale and Terms of Use and (2) consent to sharing your IP and browser information used by this site’s security protocols as outlined in our Privacy Policy.

Error Report

Sign in before submitting feedback.

Sign In

Sign In to View Your Questions

Sign In

Share This Question

Enjoying UBQ? Share the 🔗 with friends!

Link Copied!
Made By Nerd-Notes.com
Download Equation Sheet
KinematicsForces
\(\Delta x = v_i t + \frac{1}{2} at^2\)\(F = ma\)
\(v = v_i + at\)\(F_g = \frac{G m_1 m_2}{r^2}\)
\(v^2 = v_i^2 + 2a \Delta x\)\(f = \mu N\)
\(\Delta x = \frac{v_i + v}{2} t\)\(F_s =-kx\)
\(v^2 = v_f^2 \,-\, 2a \Delta x\) 
Circular MotionEnergy
\(F_c = \frac{mv^2}{r}\)\(KE = \frac{1}{2} mv^2\)
\(a_c = \frac{v^2}{r}\)\(PE = mgh\)
\(T = 2\pi \sqrt{\frac{r}{g}}\)\(KE_i + PE_i = KE_f + PE_f\)
 \(W = Fd \cos\theta\)
MomentumTorque and Rotations
\(p = mv\)\(\tau = r \cdot F \cdot \sin(\theta)\)
\(J = \Delta p\)\(I = \sum mr^2\)
\(p_i = p_f\)\(L = I \cdot \omega\)
Simple Harmonic MotionFluids
\(F = -kx\)\(P = \frac{F}{A}\)
\(T = 2\pi \sqrt{\frac{l}{g}}\)\(P_{\text{total}} = P_{\text{atm}} + \rho gh\)
\(T = 2\pi \sqrt{\frac{m}{k}}\)\(Q = Av\)
\(x(t) = A \cos(\omega t + \phi)\)\(F_b = \rho V g\)
\(a = -\omega^2 x\)\(A_1v_1 = A_2v_2\)
ConstantDescription
[katex]g[/katex]Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface
[katex]G[/katex]Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex]
[katex]\mu_k[/katex] and [katex]\mu_s[/katex]Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion.
[katex]k[/katex]Spring constant, in [katex]\text{N/m}[/katex]
[katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex]Mass of the Earth
[katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex]Mass of the Moon
[katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex]Mass of the Sun
VariableSI Unit
[katex]s[/katex] (Displacement)[katex]\text{meters (m)}[/katex]
[katex]v[/katex] (Velocity)[katex]\text{meters per second (m/s)}[/katex]
[katex]a[/katex] (Acceleration)[katex]\text{meters per second squared (m/s}^2\text{)}[/katex]
[katex]t[/katex] (Time)[katex]\text{seconds (s)}[/katex]
[katex]m[/katex] (Mass)[katex]\text{kilograms (kg)}[/katex]
VariableDerived SI Unit
[katex]F[/katex] (Force)[katex]\text{newtons (N)}[/katex]
[katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy)[katex]\text{joules (J)}[/katex]
[katex]P[/katex] (Power)[katex]\text{watts (W)}[/katex]
[katex]p[/katex] (Momentum)[katex]\text{kilogram meters per second (kgm/s)}[/katex]
[katex]\omega[/katex] (Angular Velocity)[katex]\text{radians per second (rad/s)}[/katex]
[katex]\tau[/katex] (Torque)[katex]\text{newton meters (Nm)}[/katex]
[katex]I[/katex] (Moment of Inertia)[katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex]
[katex]f[/katex] (Frequency)[katex]\text{hertz (Hz)}[/katex]

General Metric Conversion Chart

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

  1. Start with the given measurement: [katex]\text{5 km}[/katex]

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

  3. Perform the multiplication: [katex]\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}[/katex]

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

Prefix

Symbol

Power of Ten

Equivalent

Pico-

p

[katex]10^{-12}[/katex]

Nano-

n

[katex]10^{-9}[/katex]

Micro-

µ

[katex]10^{-6}[/katex]

Milli-

m

[katex]10^{-3}[/katex]

Centi-

c

[katex]10^{-2}[/katex]

Deci-

d

[katex]10^{-1}[/katex]

(Base unit)

–

[katex]10^{0}[/katex]

Deca- or Deka-

da

[katex]10^{1}[/katex]

Hecto-

h

[katex]10^{2}[/katex]

Kilo-

k

[katex]10^{3}[/katex]

Mega-

M

[katex]10^{6}[/katex]

Giga-

G

[katex]10^{9}[/katex]

Tera-

T

[katex]10^{12}[/katex]

  1. 1. Some answers may vary by 1% due to rounding.
  2. Gravity values may differ: \(9.81 \, \text{m/s}^2\) or \(10 \, \text{m/s}^2\).
  3. Variables can be written differently. For example, initial velocity (\(v_i\)) may be \(u\), and displacement (\(\Delta x\)) may be \(s\).
  4. Bookmark questions you can’t solve to revisit them later
  5. 5. Seek help if you’re stuck. The sooner you understand, the better your chances on tests.

Phy Pro

The most advanced version of Phy. 50% off, for early supporters. Prices increase soon.

$11.99

per month

Subscribe

Billed Monthly. Cancel Anytime.

Trial  –>  Phy Pro

  • Unlimited Messages and Images
  • Unlimited UBQ Credits
  • 157% Better than GPT
  • 30 --> 300 Word Input
  • 3 --> 15 MB Image Size Limit
  • 1 --> 3 Images per Message
  • All Smart Actions
  • Mobile Snaps
  • Focus Mode
  • No Ads

You can close this ad in 5 seconds.

Ads show frequently. Upgrade to Phy Pro to remove ads.

You can close this ad in 7 seconds.

Ads display every few minutes. Upgrade to Phy Pro to remove ads.

You can close this ad in 5 seconds.

Ads show frequently. Upgrade to Phy Pro to remove ads.

How to UBQ

A quick explanation

1. UBQ Credits

Credits are used to grade your FRQs and GQs. Pro users get unlimited credits. 

2. Auto Track Stats

Submitting counts as 1 attempt. 

Viewing answers or explanations count as a failed attempts.

3. Point System

Phy gives partial credit if needed

MCQs and GQs are are 1 point each. FRQs will state points for each part.

Phy helps customize your learning experience.

4. Question Explantions

Phy customizes problem explanations based on what you struggle with. Just hit the explanation button to see.

Phy gives feedback.

Understand you mistakes quicker.

4. Feedback

Phy automatically provides feedback so you can improve your responses. 

10 Free Credits To Get You Started 

Start Solving

By continuing you agree to nerd-notes.com Terms of Service, Privacy Policy, and our usage of user data. 

Jason here! Feeling uneasy about your next physics test? We will help boost your grade in just two hours.

Book a Free Lesson

NEW! PHY instantly solves any question

Try Now

NEW! Elite Members get 40% off Physics Tutoring

Join Now
We use site cookies to improve your experience. By continuing to browse on this website, you accept the use of cookies as outlined in our privacy policy.
I Accept