Slowing down of space cehicle. A space vehicle has mass 200 kg and eross-sectional area 2 ×10^4

Slowing down of space cehicle. A space vehicle has mass 200...

Slowing down of space cehicle. A space vehicle has mass $200 \mathrm{~kg}$ and eross-sectional area $2 \times 10^{4} \mathrm{~cm}^{2} .$ It travels in a region without appreciable gravitational field through a rarefied atmosphere whose mass density is $2 \times 10^{-15} \mathrm{~g} / \mathrm{ce}$ with initial speed $7.6 \times 10^{5} \mathrm{em} / \mathrm{s}$. (These would be approximate values for a satellite $500 \mathrm{~km}$ above the surface of the earth.) Assume that the conditions of the example on page 183 apply, that all the gas the space vehicle encounters sticks to it. (a) Work out the value of $c$. Consider the mass of gas picked up in $1 \mathrm{~s}$. Ans. $c=4 \times 10^{-11} \mathrm{~g} / \mathrm{cm}$. (b) Using the conservation of momentum $M v=M_{0} v_{0}$, find the differential equation for $v$ in terms of $t$ and constants. $\mathrm{Ans}, d v / d t=-c v^{3} / M_{0} v_{0^{-}}$ (c) Solve for $v$ and find the time required for the satellite to slow down to $0.9$ of its initial speed.

Slowing down of space cehicle. A space vehicle has mass $200 \mathrm{~kg}$ and eross-sectional area $2 \times 10^{4} \mathrm{~cm}^{2} .$ It travels in a region without appreciable gravitational field through a rarefied atmosphere whose mass density is $2 \times 10^{-15} \mathrm{~g} / \mathrm{ce}$ with initial speed $7.6 \times 10^{5} \mathrm{em} / \mathrm{s}$. (These would be approximate values for a satellite $500 \mathrm{~km}$ above the surface of the earth.) Assume that the conditions of the example on page 183 apply, that all the gas the space vehicle encounters sticks to it.
(a) Work out the value of $c$. Consider the mass of gas picked up in $1 \mathrm{~s}$. Ans. $c=4 \times 10^{-11} \mathrm{~g} / \mathrm{cm}$.
(b) Using the conservation of momentum $M v=M_{0} v_{0}$, find the differential equation for $v$ in terms of $t$ and constants. $\mathrm{Ans}, d v / d t=-c v^{3} / M_{0} v_{0^{-}}$
(c) Solve for $v$ and find the time required for the satellite to slow down to $0.9$ of its initial speed.

Key Concepts

Conservation of Momentum in Variable Mass Systems

This concept involves applying the law of conservation of momentum to systems where mass is not constant. When an object, such as a satellite, collects mass from its surrounding medium, its momentum is redistributed between its increasing mass and its velocity. Despite the complexity added by the changing mass, the total momentum remains conserved in the absence of external forces, forming the basis for many derivations in dynamics.

Mass Accretion and Its Effects

Mass accretion refers to the process by which an object gains mass over time through interactions with its surrounding medium. In dynamics, this process must be carefully quantified, often by relating the mass accumulated to factors such as the object’s cross-sectional area and the density of the material encountered. This directly impacts the object's velocity as the momentum conservation law implies that an increase in mass typically leads to a decrease in speed.

Differential Equations in Dynamics

The use of differential equations in dynamics is crucial for modeling systems where quantities such as mass and velocity change continuously over time. When an object experiences mass accretion, the resulting changes in momentum lead to differential equations that describe how the velocity evolves. These equations can often be solved by separation of variables, which, when combined with appropriate initial conditions, provide insights into the time evolution of the system.

Transcript

00:01 In this question we have given a uniform ladder of length 3 meter and weight 20 kilograms which rest against a frictionless wall we have to calculate the reaction force by the wall and the floor so first of all i am drawing the diagram for this given cushion so this is the frictionless wall and this is a floor and there is a ladder so this is our ladder which is resting i'm going to show the forces acting on this ladder so one force is the reaction force by this wall that is acting in this direction let this is r1 and one is the normal force by this floor that is in vertical direction this is equals to n and one is the frictional force between the ladder and this floor that is f so we can see that the resultant of this both the force is this is given by that is r2 so this is our reaction force r2 weight of this ladder is acting vertically downward weight is 20 g and this distance is given to us that is 1 meter and this distance is given to us that is length of the ladder that is 3 meter so we can say this is equal to 2 root 2 meter from pythagoras now let this point is a this is the gravity center of gravity and this is point b so we can say that for translatory equilibrium net horizontal force should be 0 so we can say that for translational equilibrium net horizontal force should be 0 so we can say that this r1 force minus the frictional force should be 0 so from here we can say that f is equals to this is r1 let this is our equation first now for we can say that vertical in vertical force should be 0 so net vertical force should also be 0 for translatory equilibrium so we can say that n should be equals to weight of the ladder that is 20 into 10 that is 200 newton this is our equation second this is the value of normal force now we can say for rotational equilibrium i'm going to take moment about center of gravity or we say take moment about point a that should be 0 so we can say that r1 into this is 2 root 2 minus moment due to the weight that is 20g into this is 1 by 2 is equal to 0 so we can say this will come out to be from here r1 and that is equals to 200 upon this is 4 root 2 so we can say this is equal to 25 root 2 newton so this is the r1 force and this is the reaction force by the wall and we can say that from equation to f should be equals to r1 that is 25 root 2 newton.

03:24 We have to calculate the value of r2 and this is equal to under root of n square plus f square.

03:31 Now put all the data in this equation.

03:33 So this is under root of 200 square plus 25 root 2 square...

Please give Ace some feedback