A 5.8 kg rock and a 2.20 x 10^-4 kg pebble are held near the surface of the eart

Question

A 5.8 kg rock and a 2.20 x 10^-4 kg pebble are held near the surface of the earth (radius = 6.38 x 10^6 m, mass = 5.98 x 10^24 kg). What is the algebraic expression for the magnitude F_rock or F_pebble of the gravitational force exerted by the earth on the rock and pebble? Express your answer in terms of the universal gravitational constant G, the mass m_earth of the earth, the mass m_rock of the rock or pebble, and the radius r_earth of the earth. (Answer using G to be the gravitational constant, m_e to be the mass of the earth, m_r to be the mass of the rock/pebble, and r to be the radius of the earth.) What are the magnitudes F_rock and F_pebble of the gravitational force exerted by the earth on the rock and pebble? What is the algebraic expression for the magnitude a_rock of the acceleration of the falling rock/pebble? Express your answer in terms of the universal gravitational constant G, the mass m_earth of the earth, and the radius r_earth of the earth. (Answer using G to be the gravitational constant, m_e to be the mass of the earth, m_r to be the mass of the rock/pebble, and r to be the radius of the earth.) What is the magnitude a_rock of the acceleration of the falling rock?

Explanation / Answer

c) F_rock = G*m_earth*m_rock/r_earth^2

d) F_rock = 6.67*10^-11*5.98*10^24*5.8/(6.38*10^6)^2

= 56.83 N

F_pebble = G*m_earth*m_pebble/r_earth^2

= 6.67*10^-11*5.98*10^24*2.2*10^-4/(6.38*10^6)^2

= 2.16*10^-3 N

e) a_rock = F_rock/m_rock

= (G*m_earth*m_rock/r_earth^2)/m_rock

= G*m_earth/r_earth^2

f) a_rock = 6.67*10^-11*5.98*10^24/(6.38*10^6)^2

= 9.8 m/s^2

a_pebble = G*m_earth/r_earth^2

= 6.67*10^-11*5.98*10^24/(6.38*10^6)^2

= 9.8 m/s^2

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