Answer:
(a)4676m/s
(b) 4227m/s
Explanation:
Let the mass of mercury be Mm = 3×10²³ kg and the mass of the object be M
Also let the distance from the center of mercury be R1 = 2.240×10⁶m let the distance the object is launched to be R
From the energy conservation equation
K1 + U1 = K2 + U2
K1, K2 = initial and final kinetic energies.
U1, U2 = initial and final potential energies.
1/2MV1² + (-GMm×M/R1) = 1/2MV2² + (-GMm×M/R)
Assuming that the very far R approaches infinity then the second term to the right side of the equation above approaches zero
So dividing through by M since M is common to both sides we have that
1/2V1² + (-G×Mm/R1) = 1/2V2²
Rearranging,
V1² = 2×( 1/2V2²+ GMm/R1 )
Given V2 = 2000m/s and also the constant value of G = 6.67×10‐¹¹ N•m²/kg²
Substituting,
V1² =2×( 1/2×2000² + 6.67×10-¹¹ ×3×10²³ /(2.24×10⁶))
V1² = 21866071.43
V1 = 4676m/s
(b) this is still similar to the case described above but this time V2 = 0m/s
V1² = 2×( 1/2V2²+ GMm/R1 ) = 2×(1/2×0+ GMm/R1) = 2GMm/R1
V1² = 2×6.67×10-¹¹ × 3×10²³/(2.24×10⁶)
= 17866071.43
V1 = 4227m/s
