Hot answers tagged acceleration
4
So you were on the right track with integrating over r and over t. Here's how you could do it:
The acceleration at any radius, r (if we assume Earth is a point mass) is:
$$a=-{GM\over r^2}$$
The minus sign is because the acceleration is anti-radial. Then you can do the following:
$$\lim_{\Delta t\rightarrow 0}~-{GM\over r^2}\Delta t~=~\Delta v$$
$$thus$$
...
4
Why don't you use energy conservation? Since this is a 1-dimensional task in potential field, it will be enough
$$
E/m = 0 - \frac{GM}{r(0)} = \frac{v(t)^2}{2} - \frac{GM}{r(t)}
$$
For your assumption that the motion is strictly radial and downwards you have $v(t) = dr(t)/dt < 0$ so you can solve for $dr(t)/dt$ and get an ordinary first order ...
4
I've always loved the Orders of magnitude articles on wikipedia. They list examples for the whole range of magnitudes for many physical quantities. There is one for acceleration too.
Among them: Rating of electronics built into military artillery shells: 15 500 G
2
If an object is moving in a circular motion, its velocity $\vec{v}$ changes. The centripetal acceleration is just a formula that gives you the length of the derivative $\frac{d\vec{v}}{dt}$ which is the acceleration. It must be caused by some force, according to Newton's second law. If you are holding the object with a rope, then it is the tension of the ...
2
The only force which works is gravity$^1$. So, change in gravitational potential energy equals final Kinetic energy(assume initial is zero).
$$mgh=mv^2/2$$
$$v=\sqrt{2gh}$$
here $h$ is vertical height traversed.See the velocity does not depend on angle of string, mass of body too..
Let's see the kinematics of body.
The length of string is $h ...
2
If $h$ is the height about the earth then
$$ \ddot{h} = -\frac{G M}{(R+h)^2} $$
$$ \ddot{h} = \frac{{\rm d} \dot{h}}{{\rm d}t}= \frac{{\rm d} \dot{h}}{{\rm d}h} \frac{{\rm d} h}{{\rm d}t} = \frac{{\rm d} \dot{h}}{{\rm d}h} \dot{h} $$
$$ \int \ddot{h}\; {\rm d} h = \int \dot{h}\; {\rm d} \dot{h} = \frac{1}{2} \dot{h}^2 + K$$
$$ \int -\frac{G M}{(R+h)^2}\; ...
1
Newton's third law doesn't imply that things can't move but it does imply conservation of momentum and energy.
Imagine a scenario where an astronaut is in orbit so they don't feel the affects of gravity. If there is an object floating and they push on it (apply a force to it) we know intuitively that the object will start to accelerate in the direction of ...
1
Your premise violates Newton's first law of motion:
If there is no net force on an object, then its velocity is constant. The object is either at rest (if its velocity is equal to zero), or it moves with constant speed in a single direction.
For an object (a body) to be accelerating there must be an external force applied. One of the reasons for ...
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