(1 point) The acceleration due to gravity, g, is given by 8= GM r2 where M...
The acceleration due to gravity, g, is constant at sea level on the Earth's surface. However, the acceleration decreases as an object moves away from the Earth's surface due to the increase in distance from the center of the Earth. Derive an expression for the acceleration due to gravity at a distance h above the surface of the Earth, 9h. Express the equation in terms of the radius R of the Earth, g, and h. 9A Suppose a 74.35 kg...
Determine the acceleration due to gravity for a point mass placed inside the Earth at a distance of r from its center
In general relativity, gravity changes time. Gravity potential o =-GM/R. Time change r= At/t= AQ / c2. If uncorrected, it would cause errors in GPS Global Positioning System. The GPS satellites orbit average 20,200 km above Earth surface, and gravity at the satellites are a little different from Earth surface. Find the: a) time error r ? e-10 due to gravity. b) distance error (km) in one day. (signal travels at light speed)
6. (10 points Extra Credit) Electrodynamics is not the only subject that utilizes Gauss' Law. We can also use it to study Newtonian gravity. The acceleration due to gravity (9can be written as, where G is Newton's gravitational constant and ρ is the m ass density. This leads us to the usual formulation of Newton's universal law of gravity,或刃--GM(f/r, as expected (if we assume V xğ-0). This "irrotational" condition allows us write (in analogy to the electric field), --Vo and...
(6 (You must show all your work). The force due to gravity on an object with mass m at a height h above the surface of the earth is _mg R F = (R+h) where is R the radius of the earth and g is the acceleration due to gravity. (a) (8 points). Express F as a series in powers of h/R. (b) (8 points). Observe that if we approximate F by the first term in the series, we get...
Acceleration of Gravity 6 Re 2R 4R Re Distance from Earth's Center 24 Earth's average sea level radius is 6.37 x 10 m, and the mass of the Earth is 5.98 x 10 kg.The gravitational constant, G = 6.67 x 10"N-m/kg . What is "g" at a distance of 2R from the center of the Earth? 2 Select one: a. 4.9 m/sec b. 1.25 m/sec? O c. zero, because you are weightless in space 2 O d. 2.45 m/sec
1. (constancy of "g") First, we ignore rotation. According to the law of gravitation the force per unit mass on an object above Earth's surface has amplitude GM/r2, wherer is the distance from the object to the center of Earth, G is the gravitational constant and M is the mass of the earth. For a point in the atmosphere at elevation z above the surface, we can write r = 2 + a where a is the radius of a...
The acceleration due to gravity at the surface of a planet is 18 m/s2. If a satellite is in a stable orbit at a distance of 3r m from the planet's center, where r is the radius of the planet, what is the speed of the satellite to maintain this stable orbit?
(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...
Acceleration due to gravity on board of your spaceship that orbits Jupiter is 25.89 m/s. How high is your spaceship above Jupiter's cloud tops? Express your answer in kilometers and round it to the nearest integer. M Jupiter = 1.90 x 104 kg; R Jupiter = 69,911 km. Your cosmic journey brought you into the vicinity of a neutron star. In fact, you are orbiting this neutron star at an orbital distance of 22.424 kilometers. The neutron star has mass...