A mass m1 of 1750 kg is fixed in place on a frictionless surface at a distance L

Question

A mass m1 of 1750 kg is fixed in place on a frictionless surface at a distance L = 10.7 mm from a second mass m2 of 189 kg. Mass m2 is free to move. The surface is slowly tilted. At what angle of inclination will the second mass begin to slide down the plane? The acceleration of gravity is 9.8 m/s2 and the value of the universal gravitational constant is 6.67259 times 10-11 N m2/kg2. Answer in units of degree A satellite moves in a circular orbit around the Earth at a speed of 5.9 km/s. Determine the satellite's altitude above the surface of the Earth. Assume the Earth is a homogeneous sphere of radius 6370 km and mass 5.98 times 1024 kg. The value of the universal gravitational constant is 6.67259 times 10-11 N m2/kg2 Answer in units of km

Explanation / Answer

1)
force between the two masses
F1 = G*m1*m2/r^2 = 0.19269 N
let theta is angle

gravitational force on m2

F2 = m2*g*sin(theta) = 1852.2*sin(theta)

here F1 = F2

0.19269 = 1852.2*sin(theta)

==> theta = sin^-1(0.19269/1852.2) = 0.006 degrees

2)

V = sqrt(G*M/(R+h))

R+h = G*M/v^2

h = G*M/v^2 - R

h = 11.458*10^6 - 6.370*10^6

h = 5.088*10^6 m = 5088 km

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