(a) The initial speed must the object be launched so that it reaches
the surface of Planet 2 with zero speed is √[2G{ M₁/ R₁ + M₂/(R₂ + L) - M₁/ (R₁ +L) - M₂/ R₂}]
(b) An inequality between M₁ and M₂ that represents when (a)
can occur is { M₁/ R₁ + M₂/(R₂ + L) - M₁/ (R₁ +L) - M₂/ R₂} ≥ 0
(c) if R₁ = R₂, then M₁ must be greater than M₂ is proved.
What is gravity?
The force of attraction felt by a person which is directed at the center of a planet or Earth is called as the gravity.
The force of attraction is directly proportional to the product of masses of the object and inversely proportional to the square of distance between them.
F = GMm/R²
Given, Planet 1 has mass M₁ and radius R₁. Planet 2 has mass M₂ and radius R₂. The two planets are a distance of L apart, measured from surface to surface. An object is launched with some initial speed from the surface of Planet 1 directly towards Planet 2. For this problem, assume that Planets1 and 2 are stationary.
If v is the launch velocity, then initial total energy will be
T.E = 1/2 mv² + ( -GM₁m/ R₁ - G M₂m/(R₂ + L)
The final total energy will be
T.E =0 + ( -GM₁m/ (R₁ +L) - G M₂m/ R₂)
From energy conservation principle, we get
1/2 mv² + ( -GM₁m/ R₁ - G M₂m/(R₂ + L) = ( -GM₁m/ (R₁ +L) - G M₂m/ R₂)
v = √[2G{ M₁/ R₁ + M₂/(R₂ + L) - M₁/ (R₁ +L) - M₂/ R₂}]
(b) an inequality between M₁ and M₂ so that object reaches the surface of Planet 2 with zero speed is
[2G{ M₁/ R₁ + M₂/(R₂ + L) - M₁/ (R₁ +L) - M₂/ R₂}] =0
{ M₁/ R₁ + M₂/(R₂ + L) - M₁/ (R₁ +L) - M₂/ R₂} ≥ 0
Thus, this is an inequality between M₁ and M₂.
(c) If R₁ = R₂, then
{ M₁/ R₁ + M₂/(R₂ + L) - M₁/ (R₁ +L) - M₂/ R₂} ≥ 0
M₁(R+L) + M₂R - M₁R -M₂(R+L) / R (R+L) ≥ 0
M₁(R+L) + M₂R - M₁R -M₂(R+L) ≥ 0
M₁L - M₂L ≥ 0
M₁ ≥ M₂
M₁ must be greater than M₂.
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