The solutions of Problems In General Physics I.E. Irodov - Universal Gravitation are given on this page. Universal Gravitation is an important topic in Physical Fundamentals of Mechanics. The force of attraction between any two objects with mass or energy is called gravitation. The main topics are Universal Gravitational Law, Strength and potential of the gravitational field of a mass point, Orbital and escape velocities, Kepler’s Law of elliptical orbit, the law of areal velocities, the law of time period, etc. Students can expect one question from this topic for the entrance exams.
Students are recommended to go through these solutions so that they can be familiar with the type of questions from universal gravitation.
1. A planet of mass M moves along a circle around the Sun with velocity v = 34.9 km/s (relative to the heliocentric reference frame). Find the period of revolution of this planet around the Sun.
Solution:
We have Mv^{2}/r = γMm_{s}/r^{2}
Or r = γm_{s}/v^{2}
Thus ω = v/r = v/γm_{s}/v^{2}
= v^{3}/γm_{s}
(Here m_{s} is the mass of the sun.)
So T = 2πγm_{s}/v^{3}
= 2π×6.67×10^{-11}×1.97×10^{30}/(34.9×10^{3})^{3}
= 1.94×10^{7} sec
= 225 days
2. The Jupiter's period of revolution around the Sun is 12 times that of the Earth. Assuming the planetary orbits to be circular, find:
(a) how many times the distance between Jupiter and the Sun exceeds that between the Earth and the Sun;
(b) the velocity and the acceleration of Jupiter in the heliocentric reference frame.
Solution:
For any planet
MRω^{2} = γMm_{s}/R^{2}
Or ω = √(γMm_{s}/R^{3})
T = 2π/ω = 2πR^{3/2}/√(γm_{s})
(a) Thus T_{J}/T_{E} = (R_{J}/R_{E})^{3/2}
R_{J}/R_{E} = (T_{J}/T_{E})^{⅔} = (12)^{2/3}
= 5.24
(b) V_{J}^{2} = γm_{s}/R_{J} and R_{J} = (T√(γm_{s})/2π)^{⅔}
So V_{J}^{2} = (γm_{s})^{⅔}(2π)^{2/3}/T^{2/3} or V_{J} = (2πγm_{s}/T)^{⅔}
Where T = 12 years. m_{s} = mass of the sun.
Putting the values we get V_{J} = 12.97 km/s
Acceleration = v_{J}^{2}/R_{J} = (2πγm_{s}/T)^{2/3}×(2π/T√(γm_{s}))^{2/3}
= (2π/T)^{4/3} (γm_{s})^{⅓}
= 2.15×10^{-4} km/s^{2}
3. A planet of mass M moves around the Sun along an ellipse so that its minimum distance from the Sun is equal to r and the maximum distance to R. Making use of Kepler's laws, find its period of revolution around the Sun.
Solution:
Semimajor axis = (r +/R)/2
It is sufficient to consider T the motion be along a circle of semi-major axis (r +/R)/2 for T does not depend on eccentricity.
Hence T = 2π((r+R)/2)^{3/2}/√(γm_{s})
= π√((r+R)^{3}/2γm_{s})
(again m_{s} is the mass of the sun)
4. A small body starts falling onto the Sun from a distance equal to the radius of the Earth's orbit. The initial velocity of the body is equal to zero in the heliocentric reference frame. Making use of Kepler's laws, find how long the body will be falling.
Solution:
We can think of the body as moving in a very elongated orbit of maximum distance R and minimum distance 0 so semi-major axis = R/2. Hence if t is the time of fall then
(2t/T)^{2} = (R/2/R)^{3} or t^{2} = T^{2}/32
Or t = T/4√2
= 365/4√2
= 64.5 days
5. Suppose we have made a model of the Solar system scaled down in the ratio η but of materials of the same mean density as the actual materials of the planets and the Sun. How will the orbital periods of revolution of planetary models change in this case?
Solution:
T = 2πR^{3/2}/√(γm_{s})
If the distances are scaled down, R^{3/2} decreases by a factor η^{3/2} and so does m_{s}. Hence T does not change.
6. A double star is a system of two stars moving around the centre of inertia of the system due to gravitation. Find the distance between the components of the double star, if its total mass equals M and the period of revolution T.
Solution:
The double star can be replaced by a single star of mass m_{1}m_{2}/m_{1} + m_{2} moving about the centre of mass subjected to the force γm_{1}m_{2}/r^{2}. Then
T = 2πr^{3/2}/√(γm_{1}m_{2}/(m_{1}m_{2}/(m_{1}+m_{2}))
= 2πr^{3/2}/√(γM)
So r^{3/2} = (T/2π)√(γM)
Or r = (T/2π)^{2/3}(γM)^{⅓}
=
7. Find the potential energy of the gravitational interaction
(a) of two mass points of masses m_{1} and m_{2} located at a distance r from each other.
(b) of a mass point of mass m and a thin uniform rod of mass M and length l, if they are located along a straight line at a distance a from each other; also find the force of their interaction.
Solution:
(a) The gravitational potential due to m_{1} at the point of location of m_{2}:
V_{2} =
= -γm_{1}/r
So U_{21} = m_{2}V_{2} = -γm_{1}m_{2}/r
Similarly U_{12} = -γm_{1}m_{2}/r
Hence U_{12} = U_{21} = U = -γm_{1}m_{2}/r
(b) Choose the location of the point mass as the origin. Then the potential energy dU of an element of mass dM = (M/l)dx of the rod in the field of the point mass is
dU = -γm(M/l)dx (1/x)
where x is the distance between the element and the point (Note that the rod and the point mass are on a straight line.) If then a is the distance of the nearer end of the rod from the point mass.
The force of attraction is
F = -∂U/∂a
= (γmM/l) ×(1/(1+l/a))(-l/a^{2})
= -γmM/a(a+l)
Minus sign means attraction.
8. A planet of mass m moves along an ellipse around the Sun so that its maximum and minimum distances from the Sun are equal to r_{1} and r_{2} respectively. Find the angular momentum M of this planet relative to the centre of the Sun.
Solution:
As the planet is under central force (gravitational interaction), its angular momentum is conserved about the Sun (which is situated at one of the foci of the ellipse.
So mv_{1}r_{1} = mv_{2}r_{2} or v_{1}^{2} = v_{2}^{2}r_{2}^{2}/r_{1}^{2} (1)
From the conservation of mechanical energy of the system (Sun + planet),
-γm_{s}m/r_{1} + ½ mv_{1}^{2} = -γm_{s}m/r_{2} + ½ mv_{2}^{2}
Or γm_{s}/r_{1} + ½ v_{2}^{2} (r_{2}^{2}/r_{1}^{2}) = -γm_{s}/r_{2} + ½ v_{2}^{2} (using (1))
Thus, v_{2} = √(2γm_{s}r_{1}/r_{2}(r_{1} + r_{2}))
Hence M = mv_{2}r_{2} = m√(2γm_{s}r_{1}r_{2}/(r_{1} + r_{2}))
9. Using the conservation laws, demonstrate that the total mechanical energy of a planet of mass m moving around the Sun along an ellipse depends only on its semi-major axis a. Find this energy as a function of a.
Solution:
From the previous problem, if r_{1}, r_{2} are the maximum and minimum distances from the sun to the planet and v_{1}, v_{2} are the corresponding velocities, then,say
E = ½ mv_{2}^{2} - γmm_{s}/r_{2}
= [γmm_{s}/(r_{1} + r_{2})](r_{1}/r_{2}) - γmm_{s}/r_{2}
= -γmm_{s}/(r_{1} + r_{2})
= γmm_{s}/2a
Where 2a = major axis = r_{1} + r_{2}.
E = ½ mr^{2} + M^{2}/2mr^{2} - γmm_{s}/r
(here M is angular momentum of the planet and m is its mass). For extreme position r = 0 and we get the quadratic
Er^{2} + γmm_{s}r - M^{2}/2m = 0
The sum of the roots of this equation are
(r_{1} + r_{2}) = -γmm_{s}/E = 2a
Thus E = -γmm_{s}/2a = constant.
10. A planet A moves along an elliptical orbit around the Sun. At the moment when it was at a distance r_{0} from the sun, its velocity was equal to v_{0} and the angle between the radius vector r_{0} and the velocity vector v_{0} was equal to α. Find the maximum and minimum distances that will separate this planet from the Sun during its orbital motion.
Solution:
From the conservation of angular momentum about the Sun.
mv_{0}r_{0} sin α = mv_{1}r_{1} = mv_{2}r_{2} or v_{1}r_{1} = v_{2}r_{2} = v_{0}r_{2} sin α (1)
From conservation of mechanical energy,
½ mv_{0}^{2} - γm_{s}m/r_{0} = ½ mv_{1}^{2} - γm_{s}m/r_{1}
Or v_{0}^{2}/2 - γm_{s}/r_{0} = v_{0}^{2}r_{0}^{2}sin^{2}α/2r_{1}^{2} - γm_{s}/r_{1} (using 1)
Or (v_{0}^{2} - 2γm_{s}/r_{0})r_{1}^{2} + 2γm_{s}r_{1} - v_{0}^{2}r_{0}^{2} sin α = 0
11. A cosmic body A moves to the Sun with velocity v_{0} (when far from the sun) and aiming parameter l the arm of the vector v_{0 }relative to the centre of the Sun (Fig. 1.51). Find the minimum distance by which this body will get to the Sun.
Solution:
At the minimum separation with the Sun, the cosmic body’s velocity is perpendicular to its position vector relative to the Sun. If r_{min} be the sought minimum distance, from conservation of angular momentum about the Sun (C).
mv_{0}l = mvr_{min} or v = v_{0}l/r_{min} (1)
From conservation of mechanical energy of the system (sun + cosmic body),
½ mv_{0}^{2} = -γm_{s}m/r_{min} + ½ mv^{2}
So v_{0}^{2}/2 = -γm_{s}/r_{min} + v_{0}^{2}/2r^{2}_{min} (using 1)
Or v_{0}^{2}r_{min}^{2} + 2γm_{s}r_{min} - v_{0}^{2}l^{2} = 0
So, r_{min} = [-2γm_{s}± √(4γ^{2}m^{2} + 4v_{0}^{2}v_{0}^{2}l^{2})]/2v_{0}^{2}
= -γm_{s} ± √(γ^{2}m_{s}^{2} + v_{0}^{4}l^{2})/v_{0}^{2}
Hence taking positive root
r_{min} = (γm_{s}/v_{0}^{2})[√(1+lv_{0}^{2}/γm_{s})^{2} - 1]
12. Demonstrate that the gravitational force acting on particle A inside a uniform spherical layer of the matter is equal to zero.
Solution:
Here we adopt a different method. Let m be the mass of the spherical layer, which is imagined to be made up of rings.
At a point inside the spherical layer at distance r from the centre, the gravitational potential due to a ring element of radius a equals
dϕ = (-γm/2ar)dl
So ϕ = ∫dϕ
= -γm/a (1)
Hence G_{r} = -dϕ/dr = 0
Hence gravitational field strength, as well as field force, becomes zero, inside a thin spherical layer.
13. A particle of mass m was transferred from the centre of the base of a uniform hemisphere of mass M and radius R into infinity. What work was performed in the process by the gravitational force exerted on the particle by the hemisphere?
Solution:
One can imagine that the uniform hemisphere is made up of thin hemispherical layers of radii ranging from 0 to R. Let us consider such a layer (Fig.). Potential at point O, due to this layer is
dϕ = -γdm/r = (-3γM/R^{3})r dr, where dm = [M/(⅔)πR^{3}](4πr^{2}/2) dr
(This is because all points of each hemispherical shell are equidistant from O.)
Hence, ϕ = ∫dϕ = (-3γM/R^{3})∫_{0}^{R} rdr = -3γM/2R
Hence, the work done by the gravitational field force on the particle of mass m, to remove it to infinity is given by the formula A = mφ , since φ = 0 at infinity.
Hence the sought work,
A_{0→∞} = -3γmM/2R
The work done by the external agent is -A.
14. Inside a uniform sphere of density ρ there is a spherical cavity whose centre is at a distance l from the centre of the sphere. Find the strength G of the gravitational field inside the cavity.
Solution:
Treating the cavity as negative mass of density - ρ and using the superposition principle, the sought field strength is :
Or
(where r_{+ }and r_{-} are the position vectors of an arbitrary point P inside the cavity with respect to centre of sphere and cavity respectively.
Thus
15. A uniform sphere has a mass M and radius R. Find the pressure p inside the sphere, caused by gravitational compression, as a function of the distance r from its centre. Evaluate p at the centre of the Earth, assuming it to be a uniform sphere.
Solution:
We partition the solid sphere into thin spherical layers and consider a layer of thickness dr lying at a distance r from the centre of the ball. Each spherical layer presses on the layers within it The considered layer is attracted to the part of the sphere lying within it (the outer part does not act on the layer).
Hence for the considered layer
dp 4πr^{2} = dF
Or dP4πr^{2} = γ(4/3)πr^{2 }drρ/r^{2}
Where ρ is the mean density of sphere.
Or dp = (4/3)πrρ^{2}r dr
Thus p = ∫_{r}^{R} dp = (2π/3)rρ^{2}(R^{2} - r^{2})
The pressure must vanish at r = R.
Or p = (⅜)(1 - (r^{2}/R^{2}))γM^{2}/πR^{4}
Putting ρ = M/(4/3)πR^{3}
Putting r = 0, we have the pressure at sphere’s centre, and treating it as earth where mean density is equal to ρ = 5.5×10^{3} kg/m^{3} and R = 64×10^{2}km
We have p = 1.73×10^{11} pa or 1.72×10^{6 }atms.
16. Find the proper potential forming energy of gravitational interaction of matter forming
(a) a thin uniform spherical layer of mass m and radius R;
(b) a uniform sphere of mass m and radius R.
Solution:
(a) Since the potential at each point of a spherical surface (shell) is constant and is equal to
ϕ = -γm/R [as we have in Eq. (1) of solution of problem 1.212]
We obtain in accordance with the equation
U = ½ ∫dm ϕ = ½ ϕ∫dm
= ½ (-γm/R)m = -γm^{2}/2R
(The factor 1/2 is needed otherwise contribution of different mass elements is counted twice]
(b) In this case the potential inside the sphere depends only on r.
ϕ = (-3γm/2R)(1 - r^{2}/3R^{2})
Here dm is the mass of an elementary spherical layer confined between the radii r and r + dr.
dm = 4πr^{2}dr ρ
= (3m/R^{3})r^{2} dr
U = ½ ∫dmϕ
= ½ ∫_{0}^{R}(3m/R^{3}) r^{2} dr{(-3γm/2R)(1 - r^{2}/3R^{2})}
After integrating, we get
U = -⅗ γ m^{2}/R
17. Two Earth's satellites move in a common plane along circular orbits. The orbital radius of one satellite r = 7000 km while that of the other satellite is ∆r = 70 km less. What time interval separates the periodic approaches of the satellites to each other over the minimum distance?
Solution:
Let ω = √(γM_{E}/r^{3}) = circular frequency of the satellite in the outer orbit.
ω_{0} = √(γM_{E}/(r - ∆r)^{3} = circular frequency of the satellite in the inner orbit.
So, relative angular velocity = ω+ω_{0}
Where - sign is to be taken when the satellites are moving in the same sense and + sign if they are moving in opposite sense. Hence, time between closest approaches
Where delta is 0 in the first case and 2 in the second case.
18. Calculate the ratios of the following accelerations: the acceleration w_{1} due to the gravitational force on the Earth's surface, the acceleration w_{2} due to the centrifugal force of inertia on the Earth's equator, and the acceleration w_{3} caused by the Sun to the bodies on the Earth.
Solution:
ω_{1} = γM/R^{2}
= 6.67×10^{-11}×5.96×10^{24}/(6.37×10^{6})^{2}
= 9.8 m/s^{2}
ω_{2} = ω^{2}R
= (2π/T)^{2}R
= (2×22/24×3600×7)^{2}6.37×10^{6}
= 0.034 m/s^{2}
And ω_{3} = γM_{s}/R^{2}_{mean}
= 6.67×10^{-11}×1.97×10^{30}/(149.5×10^{6}×10^{3})^{2}
= 5.9×10^{-3} m/s^{2}
Then ω_{1} : ω_{2} : ω_{3} = 1 : 0.0034 : 0.0006
19. At what height over the Earth's pole the free-fall acceleration decreases by one per cent; by half?
Solution:
Let h be the sought height in the first case. So
99g/100 = γM/(R+h)^{2}
= γM/R^{2}(1+h/R)^{2}
= g/(1+h/R)^{2}
Or 99/100 = (1+h/R)^{-2}
From the statement of the problem, it is obvious that in this case h<<R
Thus 99/100 = (1-2h/R) or h = R/200
= 6400/200 = 32 km
In the other case if h’ be the sought height, then
g/2 = g(1+h’/R)^{-2} or ½ = (1+h’/R)^{-2}
From the language of the problem, in this case h’ is not very small in comparison with R.
Therefore in this case we cannot use the approximation adopted in the previous case.
Here (1+h’/R)^{2} = 2
So h’/R = ±√2 - 1
As -ve sign is not acceptable
h’ = (√2 - 1)R
= (√2 - 1)6400 km
= 2650 km
20. On the pole of the Earth a body is imparted velocity v_{0} directed vertically up. Knowing the radius of the Earth and the freefall acceleration on its surface, find the height to which the body will ascend. The air drag is to be neglected.
Solution:
Let the mass of the body be m and let it go upto a height h.
From conservation of mechanical energy of the system
-γMm/R + ½ mv_{0}^{2} = -γMm/(R+h) + 0
Using γM/R^{2 }= g, in above equation and on solving we get,
h = Rv_{0}^{2}/(2gR - v_{0}^{2})