Search type Search syntax
Tags [tag]
Exact "words here"
Author user:1234
user:me (yours)
Score score:3 (3+)
score:0 (none)
Answers answers:3 (3+)
answers:0 (none)
isaccepted:yes
hasaccepted:no
inquestion:1234
Views views:250
Sections title:apples
body:"apples oranges"
URL url:"*.example.com"
Favorites infavorites:mine
infavorites:1234
Status closed:yes
duplicate:no
migrated:no
wiki:no
Types is:question
is:answer
Exclude -[tag]
-apples
For more details on advanced search visit our help page
Results tagged with Search options user 1268

The Newtonian model of gravity in which the force between two objects is given by GMm/r^2.

2
votes
Some simple scaling relations suffice to determine the size beyond which gravity prevents non-spherical rocks from forming: A molecule of mass $m$ is bound to a mass $M$ of linear size $R$ with gravi …
answered Aug 13 '14 by Johannes
1
vote
No, to translate escape velocity $v_{esc}$ at the surface of a planet into gravitational acceleration $g$ at the same location, you also need the radius $R$ of the planet. The equation that applies is …
answered Dec 7 '14 by Johannes
2
votes
The standard approach in numerical simulations is to do a discrete time stepping. If the motion is limited to the vertical direction, you just need to keep track of vertical position $z$ and vertical …
answered Aug 23 '14 by Johannes
0
votes
In broad terms, Newtonian gravity fails when gravitational accelerations become potent enough to result in relativistic speeds. In other words, Newtonian gravity is a weak gravity theory that breaks d …
answered Apr 27 '13 by Johannes
1
vote
When at distance $R$ from the center of earth, the minimum velocity needed to escape from earth's gravity is $v_{esc} = \sqrt{2GM/R}$. Here, $G$ denotes Newton's gravitational constant, and $M$ earth' …
answered Jun 4 '14 by Johannes
2
votes
Gravitational acceleration at the surface of earth varies with latitude (North-South position). This is due to 1) the outward centrifugal force produced by Earth's rotation, and 2) the equatorial bulg …
answered Dec 14 '14 by Johannes
3
votes
Calculating the gravitational force on the axis of a ring is equivalent to calculating the gravitational force of a pair of opposing small portions of the ring in which the full mass $M$ of the ring i …
answered Mar 26 '15 by Johannes
6
votes
How fast would a sphere need to rotate for a dust speck at its equator to achieve balance between gravitational attraction and centrifugal force? If you do the math (equating $G M m / R^2$ to $m \ome …
answered Sep 20 '14 by Johannes
4
votes
No. You must have made an error in determining the GR horizon radius. The radius of the event horizon determined from Newtonian theory (simply determining the distance from a point mass at which the …
answered Sep 21 '14 by Johannes
2
votes
Your starting point "normal methods of acceleration [..] add [..] constant energy over time to an object" is not correct. A constant force I would classify as "a normal method of acceleration". Such …
answered Jul 21 '13 by Johannes
5
votes
Short answer: a balloon can't float in space (above earth's atmosphere) for the same reason a ship can't float above water.
answered Nov 30 '12 by Johannes
9
votes
One can get an order of magnitude estimate of the maximum speed attainable by gravitational slingshots without doing any real calculation. The 'rough physics' reasoning goes as follows: The gravita …
answered Jul 28 '14 by Johannes
7
votes
Some rough figures: earth's mass is about $6 10^{24}$ kg. The mass of the total world population is roughly 7 billion times 80 kg or about $6 10^{11}$ kg. So earth is 13 orders of magnitude (10 trilli …
answered Jul 11 '13 by Johannes
16
votes
The gravitational binding energy for a spherical object of mass $M$ and radius $R$ is given by: $$E_{grav}=\frac35 \frac{GM^2}{R}$$ The interfacial energy for a spherical droplet is simply proportiona …
answered Mar 20 '16 by Johannes
21
votes
You would feel weightless if every part of your body of mass $m$ would be subject to an upward force equal to $m$ times the local gravitational acceleration $g$. Such an exact part-by-part cancellatio …
answered Aug 7 '13 by Johannes

15 30 50 per page