Question

a-A scientist is on a high mountain top with a gravimeter that records the free fall acceleration. The gravimeter reads that the free fall acceleration is 0.00600 m/s2 less than the free fall acceleration at sea level. What is the scientists altitude? Take the radius of the earth to be 6.37×106 m and the mass of the earth to be 5.98×1024 kg.

1.95×103 m

1.89×107 m

6.37×106 m

9.01×106 m

b-A rocket has a speed of 1.87×10⁴ m/s as it is launched from the surface of the Earth. What is its speed when it is very far away from the Earth? Take the radius of the Earth to be 6.37×10⁶ m and its mass to be 5.98×10²⁴ kg.

1.12×10⁴ m/s

2.03×10⁴ m/s

1.50×10⁴ m/s

2.18×10⁴ m/s

Answer 1

As g is proportional to 1/r^2
and g 2 = g - 0.0085
then (R2/R)^2 = (g/g2) = 9.83 / ( 9.83- 0.0060)

R2 = R * sqrt( 9.83 / ( 9.83- 0.0060) )
h = R2 - R = R * (sqrt( 9.83 / ( 9.83- 0.0060) ) -1)

= 1.944* 10 ^ 3 m altitude.

2. The conservation of energy says K2+U2 = K1+U1

The final gravitational energy, U2 is 0 because r goes to infinity.

So you have: K2=K1+U1
which is (m(vf)^2)/2 = (m(vi)^2)/2 - GMm/ R

solving for vf you get vf = sq rt [(vi^2- (2GM/R)]

G=6.67e-11 Nm^2/kg^2
M=5.98e24 kg
R=6.37e6 m

Substitute into the equation for vf, all other units cancel and your left with about 14981.900 m/s