Notice that
m has canceled out of the equation. The escape velocity is the same for all objects, regardless of mass. Also, we are not restricted to the surface of the planet;
R can be any starting point beyond the surface of the planet.
Escape from earth
What is the escape speed from the surface of Earth? Assume there is no energy loss from air resistance. Compare this to the escape speed from the Sun, starting from Earth’s orbit.
Strategy
We use
[link] , clearly defining the values of
R and
M . To escape Earth, we need the mass and radius of Earth. For escaping the Sun, we need the mass of the Sun, and the orbital distance between Earth and the Sun.
Solution
Substituting the values for Earth’s mass and radius directly into
[link] , we obtain
That is about 11 km/s or 25,000 mph. To escape the Sun, starting from Earth’s orbit, we use
$R={R}_{\text{ES}}=1.50\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{11}\phantom{\rule{0.2em}{0ex}}\text{m}$ and
${M}_{\text{Sun}}=1.99\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{30}\phantom{\rule{0.2em}{0ex}}\text{kg}$ . The result is
${v}_{\text{esc}}=4.21\phantom{\rule{0.2em}{0ex}}\times \phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{m/s}$ or about 42 km/s.
Significance
The speed needed to escape the Sun (leave the solar system) is nearly four times the escape speed from Earth’s surface. But there is help in both cases. Earth is rotating, at a speed of nearly 1.7 km/s at the equator, and we can use that velocity to help escape, or to achieve orbit. For this reason, many commercial space companies maintain launch facilities near the equator. To escape the Sun, there is even more help. Earth revolves about the Sun at a speed of approximately 30 km/s. By launching in the direction that Earth is moving, we need only an additional 12 km/s. The use of gravitational assist from other planets, essentially a gravity slingshot technique, allows space probes to reach even greater speeds. In this slingshot technique, the vehicle approaches the planet and is accelerated by the planet’s gravitational attraction. It has its greatest speed at the closest point of approach, although it decelerates in equal measure as it moves away. But relative to the planet, the vehicle’s speed far before the approach, and long after, are the same. If the directions are chosen correctly, that can result in a significant increase (or decrease if needed) in the vehicle’s speed relative to the rest of the solar system.
Visit this
website to learn more about escape velocity.
Check Your Understanding If we send a probe out of the solar system starting from Earth’s surface, do we only have to escape the Sun?
The probe must overcome both the gravitational pull of Earth and the Sun. In the second calculation of our example, we found the speed necessary to escape the Sun from a distance of Earth’s orbit, not from Earth itself. The proper way to find this value is to start with the energy equation,
[link] , in which you would include a potential energy term for both Earth and the Sun.
As stated previously, escape velocity can be defined as the initial velocity of an object that can escape the surface of a moon or planet. More generally, it is the speed at
any position such that the
total energy is zero. If the total energy is zero or greater, the object escapes. If the total energy is negative, the object cannot escape. Let’s see why that is the case.
Questions & Answers
when I click on the links in the topics noting shows. what should I do.
At time to = 0 the current to the DC motor is reverse, resulting in angular displacement of the motor shafts given by angle = (198rad/s)t - (24rad/s^2)t^2 - (2rad/s^3)t^3
At what time is the angular velocity of the motor shaft zero
In three experiments, three different horizontal forces are ap-
plied to the same block lying on the same countertop. The force
magnitudes are F1 " 12 N, F2 " 8 N, and F3 " 4 N. In each experi-
ment, the block remains stationary in spite of the applied force.
Rank the forces according to (a) the
Sadiku
Given two vectors, vector C which is 3 units, and vector D which is 5 units. If the two vectors form an angle of 45o, determine C D and direction.
Hooke's law states that the extension produced is directly proportional to the applied force provided that the elastic limit is not exceeded.
F=ke;
Shaibu
thanks
Aarti
You are welcome
Shaibu
thnx
Junaid
what is drag force
Junaid
A backward acting force that tends to resist thrust
Ian
solve:A person who weighs 720N in air is lowered in to tank of water to about chin level .He sits in a harness of negligible mass suspended from a scale that reads his apparent weight .He then dumps himself under water submerging his body .If his weight while submerged is 34.3N. find his density
The weight inside the tank is lesser due to the buoyancy force by the water displaced.
Weight of water displaced = His weight outside - his weight inside tank
= 720 - 34.3 = 685.7N
Now, the density of water = 997kg/m³ (this is a known value)
Volume of water displaced = Mass/Density
(next com)
Sharath
density or relative density
Shaibu
density
Ian
Upthrust =720-34.3=685.7N
mass of water displayed = 685.7/g
vol of water displayed = 685.7/g/997
hence, density of man = 720/g / (685.7/g/997)
=1046.6 kg/m3
1046.8
R.d=weight in air/upthrust in water
=720/34.3=20.99
R.d=density of substance/density of water
20.99=x/1
x=20.99g/cm^3
Shaibu
Kg /cubic meters
how please
Shaibu
Upthrust = 720-34.3=685.7N
vol of water = 685.7/g/density of water = 685.7/g/997
so density of man = 720/g /(685.7/g/997)
=1046.8 kg/m3
is there anyway i can see your calculations
Ian
Upthrust =720-34.3=685.7
Upthrust 720-34.3
=685.7N
Vol of water = 685.7/g/997
Hence density of man = 720/g / (685.7/g/997)
=1046.8 kg/m3
so the density of water is 997
Shaibu
Yes
Okay, thanks
Shaibu
try finding the volume then
Ian
Vol of man = vol of water displayed
I've done that; I got 0.0687m^3
Shaibu
okay i got it thanks
Ian
u welcome
Shaibu
HELLO kindly assist me on this...(MATHS) show that the function f(x)=[0 for xor=0]is continuous from the right of x->0 but not from the left of x->0
Same here, the function looks very ambiguous. please restate the question properly.
Sharath
please help me solve this problem.a hiker begins a trip by first walking 25kmSE from her car.she stops and sets her tent for the night . on the second day, she walks 40km in a direction 60°NorthofEast,at which she discovers a forest ranger's tower.find components of hiker's displacement for each day
Take a paper. put a point (name is A), now draw a line in the South east direction from A. Assume the line is 25 km long. that is the first stop (name the second point B)
From B, turn 60 degrees to the north of East and draw another line, name that C. that line is 40 km long. (contd.)
Sharath
Now, you know how to calculate displacements, I hope?
the displacement between two points is the shortest distance between the two points. go ahead and do the calculations necessary. Good luck!
Sharath
thank you so much Sharath Kumar
Liteboho
thank you, have also learned alot
Duncan
No issues at all. I love the subject and teaching it is fun. Cheers!
A projectile is thrown with a speed of v at an angle of theta has a range of R on the surface of the earth. For same v and theta,it's range on the surface of moon will be
Using some kinematics, time taken for the projectile to reach ground is (2*v*g*Sin (∆)) (here, g is gravity on Earth and ∆ is theta)
therefore, on Earth, R = 2*v²*g*Sin(∆)*Cos(∆)
on moon, the only difference is the gravity. Gravity on moon = 0.166*g
substituting that value in R, we get the new R
Sharath
Some corrections to my old post.
Time taken to reach ground = 2*v*Sin (∆)/g
R = (2*v²*Sin(∆)*Cos(∆))/g
I put the g in the numerator by mistake in my old post. apologies for that.
R on moon = (R on Earth)/(0.166)
Every body will continue in it's state of rest or of uniform motion in a straight line, unless it is compelled to change that state by an external force.
Kumaga
if you want this to become intuitive to you then you should state it
Shii
changing the state of rest or uniform motion of a body
koffi
if a body is in rest or motion it is always rest or motion, upto external force appied on it. it explains inertia
Omsai
what is a vector
smith
a ship move due north at 100kmhr----1 on a River flowing be due east on at 25kmperhr. cal the magnitude of the resultant velocity of the ship.
The result is a simple vector addition. The angle between the vectors is 90 degrees, so we can use Pythagoras theorem to get the result.
V magnitude = sqrt(100*100 + 25*25) = 103.077 km/hr.
the direction of the resultant vector can be found using trigonometry. Tan (theta) = 25/100.
An object continues to be in its state of rest or motion unless compelled by some external force
Alem
First law (law of inertia)- If a body is at rest, it would remain at rest and if the body is in the motion, it would be moving with the same velocity until or unless no external force is applied on it. If force F^=0 acceleration a^=0 or v^=0 or constant.