Weight of satellite?
distance of satellite r = R+h = 6400 km + h = 12800 hm 1+(h/R) = (r/R) = 12800/6400 = 2 h = altitude above earth ---------------- let m = mass of satellite Force of gravity F = m g = GMm/r^2 = GMm/(R+h)^2 (F/m) = [GM/R^2][1/(1+ h/R)^2] (F/m) = g (earth) [1/(1+ h/R)^2] >> here g(earth) = 9.8 m/s^2 m g(earth) = F*[(1+ h/R)^2] weight on earth = W (earth) = F*[(1+ h/R)^2] >>> (1+(h/R) =2 W = 102 *[ 2^2] = 102*4 W = 408 N
How fast would you have to go to be weightless at ground level?
So in physics class it was stated that the earth drops off at roughly 3m for every 8000m traveled. How fast would you have to go to counter gravity and essentially be in circular orbit around the surface of the earth at ground level? I tried working it out and ended up with 588mph and some thousand mph when converted. Help me answer this question plz, and join me in my random ponderings.
What is the weight of a satellite in orbit around the earth?
Plug the numbers into the Newtons Equation : F = (GMm) / r^2 F = Gravitational Force / weight G = Universal gravitational constant 6.673*10^-11 (note the - sign on the exponent) M = mass of earth (kg) 5.98*10^24 kg m = mass of satellite (kg) 150kg R = radius of earth (m) 6,378,137 m h = height above earth (m) 12,800,000m r = h+R The answer will come out as the weight being @ 162.67N Not weightless, although someone standing on the satellite would feel weightless as its in continuous free fall. Ever been in an elevator and felt lighter as it falls or been weighless on the big dipper as it plummets? The gravitational force (weight) of 162.67N is what keeps it in orbit and prevents it from flying off at a tangent. Basically it pulls the satellite towards the earth but the satellite is travelling so fast it misses and continues round in its orbit. Added: Weight is measured in Newtons and is the attractive force between the two masses. (M & m) In this case the earth and the satellite. A 1 Kg mass on the surface of the earth will be pulled towards the earth's centre with a force of 9.8N, therefore its weight at ground level is 9.8N. If you plug take m as 1kg, h as 0 and plug them into Newton's formula should get the answer 9.8N, which is where we get the constant g from. As you get further away from the centre of the mass the gravity field decreases in inverse proportion to the distance squared. Hence the divide by r^2 on Newtons fornula. Therefore the weight of a 1Kg mass at 12,800km is 1.08N. Because the satellite is in a stable orbit the opposing forces of the satellite trying to escape orbit and the gravitational pull towards the earth are equal so yes the weight is 162.7N and the centrefugal force is 162.7N. This is the correct answer to the problem. As an aside; for simplicicities case in this scenario the mass of the satellite can be regarded as constant due to its relatively low velocity compared to the speed of light. However as you come closer to the speed of light mass increases according to Einstein's theory of relativity, eventually reaching infinity at light speed.
What is the weight of a 2000 kg satellite in a geosynchronous orbit?
Not zero! Inside the satellite your *apparent* weight (if you could get inside!) is zero because you are in a state of free-fall when in orbit. But that's not the question. The satellite's weight is the gravitational attraction of the earth. This is the resultant force causing the centripetal acceleration (i.e. causing the circular motion). There is no other force on the satellite apart from its weight. Centrifugal force is what is called a fictitious force - you experience it when moving inside something moving a circle (e.g. a car going round a bend). There are several ways to work this out. This is just one... Centripetal acceleration = ω²r = (2π/T)²r The value of 'g' for an orbital satellite is the same as the centripetal acceleration (note it is less than the ground-level value of 9.8m/s² g = (2π/T)²r Geosynchronous orbits have a period of T= 1day = 24*3600s The radius of a geosynchronous orbit is 42164km = 4.2165x10⁷ m (this is the radius of the earth plus the height of the satellite above the surface). So at the height of the satellite: g = (2π/(24*3600))²* 4.2165x10⁷ = 0.223m/s² The satellite's weight = mg = 2000 x 0.223 = 446N
Height of satellite above Earth's surface?
Your method is correct. Of course, the small m's cancel out. I believe that I get the correct answer using your method. w^2*(r+h)=G*Me/(r+h)^2 where w: orbital velocity in radians per second r: earth radius Me: mass of Earth (from google) G: gravitational constant (from google) substitute w = 2*pi/T, where T is the rotation time of the Earth (for more precision, you should use the sidereal day, which is 23 hours, 56 minutes, 4 seconds -see ref 2). We can now solve for h to get the following. h = (1/2pi)*2^(1/3)*(G*Me*T^2*pi)^(1/3)-r Substituting, I get h = 3.5799*10^7 m = 22,244 miles Is this what you were expecting? It is the height above the Earth's surface for a geostationary orbit (see ref 1).
Can I mount a satellite dish near the ground?
I would try to mount it out reach, but you can mount it low. When I moved into this house I couldn't get signal due to all the trees. I finally noticed a hole in the line with where the satellite is, so I got a compass and protractor to see if feasible. Turns out that was the only place I could mount it, with a only a small window to the line of sight. It turned out I needed a post about ten feet away from the house at about 8' Any lower or higher, any other spot wouldn't work. Just too many big trees. I would try to go higher than 5-6 feet if possible. But if that's where it needs to be, low to the ground, then yeah, then that is what needs to happen. One thing is it ok with the apartment owner? And the concrete bucket idea would work. It would make moving it latter easy. Of course, it would also make stealing it or just moving it just to screw with you easy, too. Pro's and Con's. BTW, I have switched from Dish to DirecTV also. But to each their own.
Determining the weight of a satellite above earth?
If you define weight as the force of gravity on a mass, it doesn't matter whether it is in freefall or not. The force of gravity still exists. The mass is 200kg here there anywhere. So you are right to use Newton's law of universal gravitation, but you need to plug in the right values (and don't forget units!) F = G m1 m2 / R^2 G = 6.67 x 10^(-11) N m^2 / kg^2 m1 = mass of earth = 5.97 x 10^24 kg m2 = 200 kg R = 6380 km + 16,000 km = 22380 km = 2.24 x 10^7 m So F = (6.67 x 10^(-11) N m^2 / kg^2) * (200 kg) * (5.97 x 10^24) kg / (2.24 x 10^7 m)^2 F = 6.67*200*5.97/2.24^2 x 10^(-11 + 24 - 14) N The weight is F = 158.7 N To contrast, the weight on the surface of the earth would be 200 kg * 9.8 m/s^2 = 1960 N
The average atmospheric pressure at ground level is 1.0×105Pa and the Earth is a sphere of radius 6400km .?
The average atmospheric pressure at ground level is 1.0×105Pa and the Earth is a sphere of radius 6400km . Assume that the acceleration due to gravity, (= 9.8ms^-2 ) is also constant throughout the atmosphere. Estimate the total mass of air in the atmosphere. Give your answer in kg. Q2. Atmospheric pressure may be assumed to be 1.0×105Pa and the temperature constant throughout a lake. Take the density of water to be 1000kgm^-3 .In what depth of lake would the pressure at the bottom be 10 times that of normal atmospheric pressure ? Remember that, at the surface of the lake, the pressure is already one atmosphere. Give your answer in metres. Take the acceleration due to gravity to be 9.8ms^-2 .