2. It has been proposed that we could explore Mars using inflated balloons to hover just above the surface. Thebuoyancy of the atmosphere would keep the balloon aloft. The density of the Martian atmosphere is p = 0.0154kg/m³ (although this varies with temperature). Suppose we construct these balloons of a thin but tough plastichaving an area density = 5.10 g/m² (i.e., each square meter has a mass of 5.10 g). We inflate the balloonswith a very light gas whose mass we can neglect.a) If such a balloon were floating just above the surface of Mars, what forces would be acting on it? Expresseach force in terms of the balloon's radius along with the parameters given above, and calculate the necessaryradius for the balloon to be in equilibrium.b) Calculate the mass of a balloon with the radius you found in part (a).c) Suppose we release a balloon of this size on Earth, where the density of the atmosphere is 1.3 kg/m³.Would it go up, go down, or hover in place? Calculate its initial acceleration.d) Back to Mars: if we make a balloon with 5 times the radius you found in part (a), it should be able tocarry a package of instruments in addition to its own weight. How heavy can the instrument package be?

Given: Density of the atmosphere of Mars (ρM) = 0.0154 kg/m3 Surface mass density = 5.10 g/m2 Earth's atmosphere density(ρE) = 1.3 kg/m3 To find: a) Radius of the balloon. b) Mass of the balloon. c) Initial acceleration of the balloon in Earth's atmosphere. d) Mass of instrument the balloon can carry if its radius is increased 5 times as that in a). Step 2: a) To find radius of the balloon which is hovering above the surface of Mars. We know,V = 43πr3 & A =4πr2 are the volume and surface area of the spherical balloon. Since, Surface mass density = 5.10 g/m2 then mass of the balloon m = (5.10 g/m2)A = (5.10 ×10-3kg/m2)×4πr2 --(1) Thus, using the formula for the buoyant force: F = ρMVgbut F = mg is the weight of the balloonmg =ρMVgm = ρMVfrom(1)5.10 ×10-3 ×4πr2 = 0.0154×43πr3r = 0.994m Step 3: b) To calculate mass of the balloon. From (1), mass of the balloon : m=(5.10 ×10-3kg/m2)×4πr2m=(5.10 ×10-3kg/m2)×4π(0.994)2 m=0.063 kg Step 4: c) To calculate initial acceleration in Earth's atmosphere. If we release the same balloon from part a on Earth which has an atmospheric density (ρE) = 1.3 kg/m3 .Volume of the balloon with the same radius V = 43πr3=43×3.14×(0.994)3 = 4.114 m3 The net force acting on the balloon : Net Force = Buoyant force - weight ma=ρEVg-mga = (ρEVg-mg)ma = (1.3×4.114×9.8-0.063×9.8)0.063a = 822.14 m/s2 (upward) Step 5: d) Mass of instrument the balloon can carry if its radius is increased to 5 times of initial radius in a). Due to the change in radius, there will be a change in the Volume and mass of the sphere: Change in Volume: V' =43π(r')3 =43π(5r)3=43π×125r3 But we know that V = 43πr3 & A =4πr2 Therefore new volume V' = 125V =125×4.114 =514.25 m3 Change in mass: m' = (5.10 ×10-3kg/m2)×4πr'2 =(5.10 ×10-3kg/m2)×4π(5r)2 from (1) we know m = (5.10 ×10-3kg/m2)×4πr2 Therefore new mass : m' = 25m =25 ×0.063 = 1.575 kg The net force acting on the balloon along with the instrument load so that it floats: F = ρMV'gbut F = (m+m')g is the weight of the balloon(m'+mload)g =ρMV'g(m'+mload) =ρMV'mload = ρMV'-m'from(1)mload= 0.0154×514.25 - 1.575mload =6.344 kg Thus, an instrument of 6.344kg can be carried with the help of a balloon on Mars.Answers: a) Radius of the balloon on Mars = 0.994 m b) Mass of the balloon on Mars = 0.063 kg c) Initial acceleration of the balloon in Earth's atmosphere = 822.14 m/s2 d) Mass of instrument the balloon can carry if its radius is increased 5 times as that in a) = 6.344 kg

Answered: 2. It has been proposed that we could…