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SU-C ral Auth Welco The Expert TAIH Web 2.0 scientifi h Textbook Soluti €- C Secure h ps://us heexpertta.com /Take Tu alAssignment.aspx Tab on line Derivative Calc ITSDocs: Create, Co pps DN D C-St how do n Unix, how d Introduction to OOP v I Assignment ll Begin Date: 4/2/2017 3:00:00 PM Due Date: 4 7/2017 9:00:00 AM End Date: 4/10/2017 ll:59:00 PM (S%) Problem 6: In this problem you will measure the gravitational constant in a series of observational experiments, making use of Newton's law of gravitation and second law of motion as Kepler's third law of planetary motion as Wel Assignment Status Click here for 20% Part (a) Newton measured the centripetal acceleration of the moon in its orbit around Earth by comparing the force Earth exerts on detailed view the moon with the force Earth exerts on an apple. He obtained a value of a 2.69x10-5 m/s If Newton had taken the mass of Earth to be Mr 24 kg and the mean distance between the centers of Earth and the moon to be R 6.05x10 3.87: 10 m. what value would he have obtained for ME the gravitational constant, in units ofN Problem Status 20% Part b) Since measuring the centripetal acceleration of an orbiting body is rather difficult. an alternative approach is to use the Completed body's rotational period instead. Enter an expression for the gravitational constant, in terms of the distance between Earth and the moon, RNE Partial Earth's mass ME, and the moon's period of rotation around Earth. T Completed 20% Part (c) Using the sion you entered in part (b) and taking the rotational period of the moon to be T 5 days, what value Completed 3.87x108 Completed would Newton have calculated for the gravitation constant, in units of Nm kg Take ME 6.05x kg and R 20% Partial Part (d) The gravitational constant may also be calculated by analyzing the motion of a rocket. Suppose a rocket is launched Partial vertically from the surface or Earth at an initial speed ofv Its initial distance from the center of Earth is Ri. the radius of Earth. Its peak distance Completed where its speed is momentarily zero is, is Rf For s ty, ignore air resistance and Earth's rotation. Enter an expression for the gravitationa Completed constant, in terms of vi. Ri Rf and Mr 10 Completed 20% Part (e) Suppose a rocket is launched as described in part (d) with an initial speed of vi 508 m/s and attains a peak altitude of H 5x1024 kg and Ri 100 m. what is the measured value of the gravitational constant. in 12 km above the surface of Earth. Taking ME 6.05 units of N m kg Grade Summary 1A DA 047 PM 4V6/2017

Explanation / Answer

b)

Centripetal acceleration, a = v^2/R

Now, v = orbital speed = 2*pi*R/T <-------- T = orbital period

So, a = (2*pi*R/T)^2/R

or , a = 4*pi^2*R/T^2

Also, the force of gravity will provide the necessary centripetal force

So, GMm/R^2 = m*a = m*(4*pi^2*R/T^2)

So, GM/R^2 = 4*pi^2*R/T^2

So, G = 4*pi^2*R^3/(T^2*M) <---------answer

where R = RME , M = ME

c)

So, G = 4*pi^2*(3.87*10^8)^3/((27.5*24*3600)^2*6.05*10^24)

So, G = 6.699*10^-11 N.m2/kg2

d)

By work energy principle,

All the KE at the start of the flight will be converted the work done against gravity

Now, work done against gravity = change in PE

So, KE = change in PE

So, 0.5*mvi^2 = GMm/Ri - GMm/Rf = GMm*(Rf - Ri)/(Ri*Rf)

here M = ME

Rf = Ri + H

So, 0.5*vi^2 = GM*(Rf - Ri)/(Ri*Rf)

So, G = 0.5*vi^2*(Ri*Rf)/((Rf - Ri)*M) <-------answer

e)

Pluggin in the values,

G = 0.5*(508)^2*(6.3*10^6)*(6.3*10^6 + 12*10^3)/(6.05*10^24*(12*10^3))

= 7.07*10^-11 N.m2/kg^2

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