earth is r.-6.38% 106 m and the gravitational constant G-667x 10-11 N m4kg2 and the Earth's...
A satellite m = 500 kg orbits the earth at a distance d = 218 km, above the surface of the planet. The radius of the earth is re = 6.38 × 106 m and the gravitational constant G = 6.67 × 10-11 N m2/kg2 and the Earth's mass is me = 5.98 × 1024 kg. What is the speed of the satellite in m/s?
(a) Explain the relationship between the universal gravitational constant G and the acceleration due to gravity at the earth's surface g. Therefore, calculate g from G using the relationships given below. Justify the choice of units for G (Nm kg?). F= mg The mass of the earth is 5.98 x 1024 kg, it's radius is Tg = 6.38 x 10 m, and G = 6.67 x 10-11 Nm?kg (10 marks) (b) Explain, including performing the integral, how the work done...
A satellite used in a cellular telephone network has a mass of 2380 kg and is in a circular orbit at a height of 850 km above the surface of the earth. Part A What is the gravitational force Fgrav on the satellite? Take the gravitational constant to be G = 6.67×10−11 N⋅m2/kg2 , the mass of the earth to be me = 5.97×1024 kg , and the radius of the Earth to be re = 6.38×106 m
i got to the end but couldnt find the right value for R. and explanation would be appreciated. 26. A satellite with mass 10 kg has a circular orbit 106 m above the Earth's surface (the radius of the Earth is 6.4 x 106 m). What is the total (kinetic +potential) energy of the satellite? Take the potential energy to be zero when the Earth and satellite are infinitely far apart. It might help to know that the mass of...
Draw a diagram and please answer both parts What was the speed of a space shuttle that orbited Earth at an altitude of 3 x 105 m? Mass of earth = 5.98 x 1024 kg; Radius of earth = 6.38 x 106 m Universal gravitational constant, G = 6.67 x 10-11 N m2 / kg (b) What is its period?
5. Use your closest value to calculate a value of the universal gravitational constant G, where the mass of the Earth Me 5.983 X 1024 kg and its radius is r=6.378 X 106 m. ME em. 6.37 10 W UESTIONS Which results gave the more accurate readings? Why is this so?
A satellite used in a cellular telephone network has a mass of 2050 kg and is in a circular orbit at a height of 880 km above the surface of the earth. Part A What is the gravitational force Fgrav on the satellite? Take the gravitational constant to be G = 6.67×10−11 N⋅m2/kg2 , the mass of the earth to be me = 5.97×1024 kg , and the radius of the Earth to be re = 6.38×106 m . part...
Earth Mass/Radius: 1 M= 5.97 x 10^24 kg 1 R= 6.38 x 10^6 m Compare th centripetal force of s 75 kg person standing on the equator of the earth to that of the gravitational force due to the earths rotation. In other words, compute the ratio ( F gravitational/ F centripetal). How many revolutions per day would the earth have to turn to make these forces equal to one another (to makes this ratio exactly 1)?
The value of g at the Earth's surface is about 9.81 m/s2. The the mass of the Earth = 5.98 x 1024 kg and the radius of the Earth is 6.37 x 106 m. 1) Calculate the value of g for a satellite that is orbiting in a circle at an altitude equal to 3 Earth radii. g =
Task 2 (Gravitational Potential of the Earth) In good approximation, the earth can be regarded as a sphere of homogenous mass density with radius R and total mass M. The gravitational potential of a mass m which has a distance r to the center of the earth is given by GMm , r>R U(r) 2R where Eo and G are constants a) Calculate the force F(r) which acts on a mass m at the earth's surface Hint: The gradient of...