HELP!! Hard problem!! Determine the satellite’s altitude above the surface of the Earth.?
A satellite moves in a circular orbit meaning that gravity is balanced by centrifugal force, or in another words, gravitational and centrifugal accelerations have opposite directions and equal magnitudes: ag = acf where - acf is centrifugal acceleration acf = v^2 / r - ag is gravitational acceleration at distance r from center of the Earth; this shouldn't be confused with g which is gravitational acceleration on the surface. ag = g for r=R, where R is radius of the Earth, ag = G * M / r^2 M is mass of the Earth, M=5.98 × 10^24 kg. G is universal gravitational constant; G= 6.67259 × 10^−11 N · m^2/kg^2 G M / r^2 = v^2 / r G M = v^2 r r = G M / v^2 r = 6.67259 * 10^−11 * 5.98 * 10^24 / 5900^2 r = 11,462,823 m = 11463 km Notice that this is radius of satellite's orbit, i.e. it's distance from center of the Earth. Its altitude above the surface is: H = r - R = 11463 - 6370 = 5093 km
What is the escape velocity from the Earth’s surface to the height of 6400 km?
There is no escape velocity to a particular distance. You can get to that distance as fast or as slowly as you like. Escape velocity is the speed at which an object must travel away from the Earth for the deceleration caused by gravity to be insufficient to stop it. It is the same speed as a body reaches when falling from an infinite distance.
Calculating orbital speed?
It's hard to tell what you're supposed to do, since I don't know which equations you've got on hand, but you really should know these: The force between two masses m and M, separated by a distance r, is: F = GMm / r² The acceleration on a body moving with constant speed v in a circle of radius r is: a = v² / r You should also know Newton's 2nd law which, for the small mass m, says: F = ma a = F / m The force is the force of gravity given above, and the acceleration produced is a circular orbit with speed v. So, substitute for a on the left v² / r = F / m ...and for F on the right: v² / r = (GMm / r²) / m = GM / r² Then solve: v² = GM / r v = sqrt(GM / r) If R is the radius of the earth, then find problem (b) by letting r = 2R, and (c) by letting r = 3R. (b) v = sqrt( GM / (2R) ) (c) v = sqrt( GM / (3R) ) Look up the mass M of the Earth, the universal gravitation constant G, and the earth's radius R. Convert to SI units, if necessary, and calculate.
If the parking orbit is 180 km above Earth's surface, what is the escape speed from the orbit?
A space orbit probe can be either launched at escape speed from Earth or transported to a "parking orbit" above Earth and then launched. I am having so much trouble with this!
Why our satellite need escape velocity as much at 28,080km/h, when Earth gravity are less outside atmosphere and Earth rotating at 1656km/h?
Hi guys, i want to know how much actual Earth gravity at outside Earth atmosphere. We all knows that Earth normal gravity inside Earth atmosphere is 9.8 meters/second^(2*time). but if we talk about gravity outside Earth atmosphere, it should be less than that, am i right? when we go up until there's no air, it means that gravity outside there should be are so weak. even gases out there is nothing left because less gravity, isn't it? back to my previous question, if my satellite successfully going up with drone until there's no air left on edge of Earth atmosphere, don't you think that my satellite should be already have escape velocity at 1656km/h due to Earth rotation, and if i moving my satellite with additional rocket made by fireworks stuff like i mention before to the reverse of direction of Earth rotation, don't you think that it's enough to make my satellite doesn't falls to Earth with let say 1800km/h escape velocity (Earth rotation + my rocket velocity) and keep it rotating on Earth orbit up there? Thanks in advance..
Why do we have to launch a satellite with escape velocity?
First of all, only inter-planetary missions are given the escape velocity. I will try to answer this as simple I can.Now, look at the above picture. Imagine you've climbed a tower or a tall building. Try to throw a ball horizontally. It will hit the ground at some point. Next, you try to throw the ball with more effort. You'll notice that it has hit the ground further away from the first point. If you try to increase the velocity continuously in subsequent attempts, eventually you will see that the ball has completed one circle and it never touches the ground. The speed at which this phenomenon has occurred is the orbital velocity at that altitude. As you go high, i.e away from the surface, one will notice that the velocity to orbit will drop inversely proportional to the square root of the distance from the center of the Earth. This the principle with which rockets are launched. Scientists calculate what is the optimum height to keep the satellite in the orbit considering many constraints. Now regarding the escape velocity, it is the minimum velocity with which an object has to be thrown so that it escapes the clutches of Earth's gravity. If you look at the above picture, the path shown by 'E' is the path taken by the object, if you throw it with an escape velocity. If you throw with velocity more than the orbital velocity at that altitude, you'll notice the object to form an ellipse about the earth. In this picture it is shown by the path 'D'. As you increase velocity, you'll notice that the ellipse is getting bigger. Eventually, there will be a velocity at which the object never returns back. This is called the escape velocity at that altitude. Suppose if your satellite is intended to go to Mars or Jupiter, then only you'll impart escape velocity. For normal satellites, escape velocity is not given. We want these satellites to continuously orbit the Earth. One more thing, if you see the above picture, paths 'C' and 'D' are ellipses about the earth whereas path 'E' is a parabola. If the satellite takes greater velocity than the escape velocity, path taken by the satellite is a hyperbola. ONCE AGAIN I WOULD LIKE TO REMIND YOU THAT ONLY IN INTER-PLANETARY MISSIONS, ESCAPE VELOCITY IS IMPARTED.
How many more times is escape speed than orbital speed for a satellite revolving near Earth?
Earth's escape velocity is 11.20 km per second. This is only slightly more than the orbital velocity of satellites orbiting Earth.Satellites that are further away from Earth orbit at lower speeds than those that are closer . The ISS at an altitude of ~400 km is orbiting Earth at 7.66 km per second whereas satellites in geosynchronous orbit at 42,000 km move at only 3 km per second.Orbital Speed
An artificial Earth satellite is “parked” in an equatorial circular orbit at an altitude of 103^km.?
what is "103^km" supposed to mean ?? Later on you use 10^3km which I assume is 1000 km. Satellite motion, circular V = √(GM/R) V = velocity in m/s G = 6.673e-11 Nm²/kg² M is mass of central body in kg R is radius of orbit in m use the above to calculate the velocity in the orbit. note that R = 1000 km plus the radius of earth, in meters Escape speed V₀ = √(2GM/R) G = 6.673e-11 Nm²/kg² where M is mass of body and R is the distance from the center of the body. Calculate V₀ from above. Subtract V from the first equation to get added speed needed.
How can we compare the escape velocity to the velocity of a satellite revolving around the Earth very close to its surface?
Let G be the gravitational force of earth,M be it’s mass and R be the radii of earth.We know that Vescape=√2(GM)/RAlso Vorbital=√(GM)/r ,where r is the distance between the centre of earth and sattelite but the sattelite is close to the surface so R~r.We can write above equation as,Vorbital=√(GM)/RVe=√2*(Vo)So we can say that escape velocity =√2 times Vo
Physics Q!! Suppose a satellite is in a circular orbit 7.600 Earth radii above the surface?
The satellite's PE in orbit = Earth's grav. potential (Ve in J/kg) x Ms (kg) Ve = -G.Me / Ro .. (Ro = orbit radius = 8.6 Re, Me = mass of Earth) Satellite PE = -G.Me.Ms / Ro (J) .. The satellite has to be given (+G.Me.Ms / Ro ----[♦]) KE to remove it completely from the Earth's grav. attraction ( to infinity, where it's net energy is zero) The satellite already has some of this KE due to it's orbit velocity (Vo) Satellite centripetal force (Ms.Vo² / Ro) = grav.attraction (G.Me.Ms / Ro²) Ms.Vo² = G.Me.Ms / Ro Orbit KE = ½ Ms.Vo² = ½G.Me.Ms / Ro ----[♠] Additional KE required = (+G.Me.Ms / Ro [♦]) - (½G.Me.Ms / Ro [♠]) = ½G.Me.Ms / Ro Additional KE = ½ Ms.Vx² .. .. Vx = escape velocity ½ Ms.Vx² = ½G.Me.Ms / Ro Vx² = G.Me / Ro Vx² = (6.67^-11 x 5.97^24kg) / (8.60 x 6.40^6m) = 7.235^6 (m/s)².. .. ►Vx = 2690.0 m/s