anonymous
  • anonymous
an object of mass 'm' is located a distance 'R' from the center of a planet with mass 'M.' The acce leration of gravity of the mass 'm' at this location is24m/s^2. Suppose the the mass of the object is halved to '0.5m' and the seperation distance is tripled to '3R' The new acceleration of gravity value for the mass and the new location would be...?
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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ash2326
  • ash2326
we know Gravitational force is given by the relation F= G*m*M/R^2 here m is the mass of small body M= mass of the planet G= universal gravitational constant R= distance of the object from the center of the planet Acceleration is given as 24 m/s^2 Force= ma=24m Newtons we are given \[1st.....24m= G*m*M/r^{2}\] second case mass is halved so it's 0.5m separation distance is tripled , so it's 3R now \[2nd.... F = G*0.5 m*M/(3R)^{2}\] dividing 1st by 2nd \[ 24m/F=G * m* M * 9r^2/(G * 0.5 m* M * r^2)\] we'll get \[ 24m/ F= 9/0.5 \] or \[ 24m/F=18 \] or \[ F= 24m/18 \] F= 0.5 m * a1 a1= new acceleration so \[ a1*0.5m= 24m/18 \] \[ a1= 48/18 \] a1= 2.666 m/s^2 so the acceleration after halving the mass , and tripling the distance is 2.666 m/s^2
anonymous
  • anonymous
that's the answer I got as well. for my working it's in the math group question you posted :)

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