Matric Notes Physics 9th Ch 5 Gravitation Long Questions
Q1. State and explain Newton’s law of gravitation?
Ans. Newton’s of universal gravitation :- This law states that in this universe Avery two bodies attract each other with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance b/w them .
Explanation:- consider two bodies a and b as shown in figure let their masses are m1 and m2 respectively and the distance b/w them is “r” then according to Newton’s law of gravitation the force of attraction :f: is given as:-
F φ m1m2 ................................. (i)
F φ 1/r2 .................(ii)
Combining eq (i) and eq (ii) we get.
F φ m1m2 1/r2
F = G m1m2 1/r2................(iii)
Eq (iii) is the mathematical form of Newton’s law of gravitation. Where “G” is gravitational constant and its value is equal to 6.673 x 10-11= N.m2/kg2
Q.2 Using law of universal gravitation find mass of earth?
Ans. We can find mass of earth “Me” by using universal gravitation by following method suppose we have a body of mass “m” laying on the on the surface m of earth as shown in figure.
Let “m” is the mass of body “Me” is the mass of earth “Re” is the radius of earth which is the distance b/w the centers of body and earth. Now according to law of universal gravitation the force b/w earth and object is given by:-
F = GMem/Re2 ...................(i)
We also know that the force with which a body is attracted toward the center of earth is equal to its weight
ie F = W = mg ..........(ii)
Combining eq (i) and eq (ii) we get.
GMem/Re2 = mg
=> GMe = gRe2 => Me = gRe2 /G ...... (iii)
Now as we know that g = 10m/sec2
Re = 6.4 x 106m G = 6.67 x 10-11 NM /kg2
Putting these values in eq (iii) we get
Me = (10) (6.4 x 10(6))2/6.67 x 10(-11)
Me = (10) (40.96) (10(12)) / 6.67 x 10(-11)
Me = 409.6 x 10(12) x 10(11) /6.67
Me = 61 x 10(23) => Me = 6 x 10(24) kg
Q.3 Explain variation of “g” with altitude?
Ans. Considers an object laying on the surface of earth as shown in figure. Where me is mass of earth Re is radius of earth and “h” is the height of object from the surface of earth then according to law of universal gravitation the formula of mass of earth is given by
Me = gRe2/G => Me G = gRe2 =>g = Me G/Re2................(i)
Eq (i) shows the formula with the help of which we can find value of “g” now according to figure we want to find value of “g” at point “A” i.e when body is laying at the surface of the earth where value of “g” is “go” and given by:=
Go = GMe/Re2................ (ii)
Now we find out value of “g” again when body is at height “h” from the surface of earth and the distance b/w their centers is equal to “Re+ h” in figure the value of “g” at point “B” is shown by “g” which is given by:-
G h = GMe / (Re + h)2................(iii)
So, eq (ii) and eq (iii) shows that the value of “g” is inversely proportion to the square of the distance from earth’s center ie the value of “g” decreases with altitude. That’s why the value of “g” is greater at poles than the equators. Similarly the value of “g” will be greater in plane areas as compare to hilly areas. For example value of “g” at Karachi will be greater than the value of “g” at Murree.
Q4. What is satellite? Derive the formula of orbital speed of an arbitral speed?
Ans. Satellite:- A satellite is an object which can move around a planet. Suppose a satellite of mass “m” is revolving around the earth in a circular orbit of radius “r” as shown then according to Newton’s law of gravitation the force of attraction b/w them is given by following:-
F = (GMem)/r2........... (1)
As the satellite performs circular motion so, the gravitation force in this acts as centripetal fore.
I.e F = (mv2)/r.............. (2)
Comparing eq(i) and eq (ii) we get,
(mv2)/r = (GMem)/r2
v2= (G Me)/r
√v2 = √(G Me)/r
=> v = √(G Me)/r ....(3)
As “r = Re +h” so, eq (iii) becomes
V = √(GMe)/(Re + h).......................(4)
Eq (4) is the formula for orbital speed of satellite.
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