Gravitational Intensity, Potential, Energy Numericals: Planets and Satellites ISC Class 11 Physics Nootan Solutions Ch-12. Step by step Solutions of Kumar and Mittal Physics of Nageen Prakashan as council latest prescribe guideline for upcoming exam. Visit official Website CISCE for detail information about ISC Board Class-11 Physics.
Gravitational Intensity, Potential, Energy Numericals: Planets and Satellites ISC Class 11 Physics Nootan Ch-12
| Board | ISC |
| Class | 11 |
| Subject | Physics |
| Writer | Kumar and Mittal |
| Publication | Nageen Prakashan |
| Chapter-12 | Gravitation: Planets and Satellites |
| Topics | Numericals on Gravitational Intensity, Gravitational Potential and Gravitational Energy |
| Academic Session | 2024-2025 |
Numericals on Gravitational Intensity, Gravitational Potential and Gravitational Energy
Ch-12 Gravitation: Planets and Satellites ISC Class 11 Physics Nootan Solutions
Que-12: The radius of the earth is 6.4 x 10^6 m and the mean density is
5.5 x 10^3 kg/m^3. Determine the gravitational potential at the earth’s surface. (G = 6.67 x 10^-11 Nm^2 kg^-2).
Solution- Given, radius of the earth, R=6.4×10^6m
Density, ρ=5.5×10^3 kg/m^3, G=6.67×10^−11Nm^2 kg^−2
∵ Mass of the earth M = volume × density =(4/3)πR^3 ρ
∴ Gravitational potential on the earth’s surface,
V=−GMR=−GR×(4/3)πR^3 ρ=−(4/3)πGR^2 ρ
=−4/3×3.14×6.67×10^−11×(6.37×10^6)^2×5.5×10^3
⇒V=−6.3×10^7 J/kg. Ans.
Que-13: Two bodies of masses 100 kg and 100 kg are at a distance 1.00 m apart. Calculate the gravitational field intensity and the potential at the middle-point of the line joining them. (G = 6.67 x 10^-11 Nm^2 kg^-2).
Solution- E = E1 + E2
E = E1-E2 = {G (1000)/(0.5)²} – (G (100)/(0.5)²}
={100G/(0.5)²} (10-1)
(900G × 100)/25
= (90000 x 6.67×10^-11)/25
= 2.4×10^-7 N/kg Ans.
M1 = 100kg
M2 = 1000kg
r = 1 m
Gravitational potential at X = (GM1)/(г/2) – (GM2)/(г/2)
(2G/r) (M1 + M2)
= (2×6.67 x 10^-11) (100+1000)
= -1.46 x 10^-7 J/kg Ans.
Que-14: The mass of the earth is 6.0 x 10^24 kg. Calculate, with sign (i) the potential energy of a body of mass 33.5 kg and (ii) the gravitational potential at a distance of 3.35 x 10^10 m from the centre of the earth
(G = 6.67 x 10^-11 SI units).
Solution- (i) U = (GMem)/Re
= (6.67 x 10^-11 x 6 x 10^24 x 33.5)/(6.4 x 10^6)
= -4.02 x 10^5 J Ans.
(ii) v =-Gme/r
= (6.7 x 10^-11 x 6.4 x 10^6)/(3.35 x 10^10)
= -12 x 10^3 J/kg Ans.
Que-15: The mass and the radius of the earth are 6.0 x 10^24 kg and 6400 km. What will be the potential energy of a 200 kg body placed 600 km high from earth’s surface ? (G = 6.67 x 10^-11 SI units).
Solution- m1 (mass of the Earth) = 6.0 × 10^24 kg
m2 (mass of the body) = 200 kg
r (distance) = Earth’s radius (6400 km) + height (600 km) = 7000 km
Now, calculate the potential energy:
PE = – (6.7 × 10^-11 N m²/kg² * (6.0 × 10^24 kg * 200 kg)) / 7,000 km
PE ≈ – (8040 × 10^10 N m²/kg²) / 7,000 km
PE ≈ – 1.15 × 10^10 Joules Ans.
Que-16: How much energy would be needed for a 100 kg body to escape from the earth ? (g = 10m/s^2 and radius of the earth Re = 6.4 x 10^6m).
Solution- Energy Required = ½ mve²
= 1/2 m(√2gR)^2
= 1/2 × 100 × 2 × 6.4 x 10^6 x 10
= 6.4×10^9 J Ans.
Que-17: A 500-kg artificial satellite is revolving around the earth at a height of 1800 km from the earth. Find out the potential energy, kinetic energy and total energy of the satellite. The earth’s radius is 6400 km and
g = 10m/s^2
Solution- U = (-GMm)/(R+h)
= {-(6.67 x 10^-11)x(6.0 x 10^24)x(500)}/(6400+1800) x 10^3
= -2.5 x 10^10 J Ans.
= K.E. = U/2 = 1.25 X 10^10 J Ans.
= T.E. = -U/2 = -1.25 X 10^10 J Ans.
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