# Gravitation : Satellites and Escape Velocity Numerical ISC Class 11 Physics Nootan

Gravitation : Satellites and Escape Velocity Numerical 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.

## Gravitation : Satellites and Escape Velocity Numerical ISC Class 11 Physics Nootan Solutions Ch-12

 Board ISC Class 11 Subject Physics Writer Kumar and Mittal Publication Nageen Prakashan Chapter-12 Gravitation: Planets and Satellites Topics Numericals on Satellite and Escape Velocity Academic Session 2024-2025

### Numericals on Satellite and Escape Velocity

Chapter-12 Gravitation: Planets and Satellites ISC Class 11 Nootan Solutions Kumar and Mittal Physics of Nageen Prakashan

##### Que-18: The distance of the moon from the earth is 3.8 x 10^5 km. Calculate the speed of the moon revolving around the earth . Mass of the earth = 6.1 x 10^24 kg and G = 6.7 X 10^-11 Nm^2 kg^-2.

Solution- V = √(GM/r)
r = R+h
V = √(GMe/R)
V = √{(6.6 x 10^-11 x 6 x 10^4)/(3.84 x 10^6)}
V = 1.03 km/s Ans.

##### Que-19: A satellite which is at a height h from earth’s surface completes one revolution of the earth in 90 minutes. If the radius of the earth be 6370 km and acceleration due to gravity 9.8 m/s^2, then find the value of h.

Solution-  R = radius of earth = 6370 km = 6370,000 m
h = height of satellite above the earth’s surface
g = acceleration due to gravity = 9.8 m/s²
T = time period of the satellite for one revolution around earth = 90 min = 90 x 60 sec = 5400 sec
time period of the satellite for one revolution around earth is given as
T = (2π) √(R + h)³/(gR²)
taking square both side
T² = (2π)² (R + h)³/(gR²)
T² (gR²)/ (2π)² = (R + h)³
inserting the values
(5400)² ((9.8) (6370,000)²)/ (2(3.14))² = ((6370,000) + h)³
h = 279532 m = 279.532 km = 280 km Ans.

##### Que-20: Calculate the orbital velocity of a satellite revolving near the earth, if the radius of the earth is 6.4 x 10^6 m and acceleration due to gravity 10 m/s^2. What will be the orbital velocity if the satellite be at a height of 2000 km from the earth’s surface?

Solution-  Re = 6.4 X 10^6 m,  g = 10 m/s²
h = 2000 km = 2000 x 1000 = 2×10^6 m
Vo = √GM/Re = √gRe √10×6.4×10^6
√64X10^6 = 8000 m/s = 8 km/s Ans.

Orbital velocity at height 2×10^6 m,
= √Gm/Re = √gRe^2/(Re+h) = √{10x(6.4×10^6)^2}/{(6.4×10^6)+(2×10^6)}
= √(10 x 6.4×10^6 x 6.4 x 10^6)/(8.4 x 10^6) = √(64 x 64 x 10^6)/84
=√48.76 x 10^3 = 6.98 km/s = 7 km/s Ans.

##### Que-21: A 500 kg satellite revolves around the earth at a height of 10^3 km above the earth’s surface. Calculate, assuming circular orbit : (i) speed of satellite, (ii) angular velocity of satellite, (iii) gravitational force of the earth on the satellite. Given : radius of the earth 6.4 x 10^6 m, g = 9.80 m/s²

Solution-  Given mass = 500 kg, g = 9.8, radius of earth = 6.4 x 10⁶ m,

(i) vₒ = √gR²/R+h = √(9.8 x 6.4² x 10⁶)/(7.4 x 10⁶) = 7.36 km/s = 7.36 x 10³ m/s. Ans.

(ii) v = wr, so w = v/r
= 7.3 x 10³/(6.4+1)x10^6

(iii) We know that force of attraction = GMm/(R+h)²

and g = GM/R² so force of attraction will be = gR²m/(R+h)²

= (9.8 x 6.4² x 10^6 x 500)/7400 = 0.00366 x 10⁶ N = 3.66 x 10³ N. Ans.

##### Que-22:  A satellite is revolving around the earth in a circular orbit of radius 8000 km. Which With what speed this satellite be projected in the orbit ? What will be its period of revolution ? (9 = 9.8 m/s^2, radius of the earth 6400 km).

Solution-  Suppose, the speed of the satellite is v. The acceleration of the satellite is v2r, where r is the radius of the orbit. The force on the satellite is GMmr2 with usual symbols. Using Newton’s second law,

GMm/r^2=mv^2/r

or, v^2 = GM/r = gR^2/r = (9.8m/s^2)(6400km)^2 /(8000km)

giving v = 7.08 km/s Ans.

The time period is 2πr/v = 2π(8000km)/(7.08km/s)≈118minutes Ans.

##### Que-23: An earth-satellite is revolving at a height of 1800 km of from the earth’s surface. Radius of the earth is 6300 km and acceleration due to gravity at the earth’s surface is 10 m/s^2. Find out : (i) orbital velocity of the satellite, (ii) radial acceleration of the satellite, (iii) period of revolution of the satellite.

Solution- (i)  V = √Gm/r = √4×10^14/(6.400+1.800)×10^6
= √4×10^8/8.200 = 2×10^4/ 2√2 = 0.707×10^4
=7.07×10^3 m/s = 7 km/s Ans.

(ii) v^2/r = GM/r^2
= gR^2/r^2
=10×40×10^12/65×10^12 = 80/13
=6m/s^2. Ans

(iii)  The period (T) of revolution of the satellite can be calculated using the formula:

T = 2πR/v

Plug in the values:

T = (2π x 1.8 x 10^6)/ 6

T = 7267 sec

So, the period of revolution of the satellite is approximately 7267 seconds  Ans.

##### Que-24: An artificial satellite is revolving at a height of 500 km above the earth’s surface in a circular orbit, completing one revolution in 98 minutes. Calculate the mass of the the earth. Given : G = 6.67 x 10^-11 Nm^2 kg^-2, radius of the earth = 6.37 x 10^6 m.

Solution-  According to the problem the satellite is completing one revolution of the circular orbital in 98 min
Now the let the velocity of the satellite is v
v = 2π(r+h)/t
Therefore the force can be calculated as
f = g Mm/(r+h)^2 = mv^2/(r+h)
= gM/(r+h)^2 =  (2π(r+h))^2/t^2(r+h)
= M = 4π^2(r+h)^2/Gt^2
now by putting all the values here
we will get M = 5.62 x 10^24kg Ans.

##### Que-25: The moon completes one revolution around the earth in 27 days in an orbit of radius 3.8 x 10^5 km. The earth completes one revolution around the sun in 365 days in an orbit of radius 1.5 x 10^8 km. Compare the masses of the sun and the earth.

Solution- Mass of earth
Me = (4π^2 x rm^3)/(Tm^2 X G)
where r is orbit of moon and T all time period
and mass of sun
Ms = (4π^2 x re^3)/(Te^2 X G)
where re is orbit of earth and Te is time period of earth
= Ms/Me = (re^3/rm^3) x (Tm^2/Te^2)
= (1.5 x 10^8)^3/(3.8 x 10^5)^3 x (27/365)^2
= 3.366 x 10^5 : 1 Ans.

##### Que-26: Taking moon’s period of revolution around the earth as 30 days (and neglecting the effect of sun and other planets on its motion), calculate its distance r from the earth (G = 6.67 x 10^-11 Nm^2 kg^-2, mass of the earth Me = 6 x 10^24 kg).

Solution-    T^2= 4π^2 R^3/GM
Where T is the time period of revolution, R is the radius of moon’s orbit, M is the mass of the planet (Earth), G is the universal gravitational constant.
We have,
T = 30 days = 2592000 s
G = 6.67 X 10-11 N(m/kg)2
M = 5.97 X 10-24 kg
Therefore,
2592000^2= 4*3.14^2 *R^3/6.67*10^-11 * 5.97*10^24
R=4.8*10^5 Ans.

##### Que-27: The density of a planet is 8 x 10^3 kg/m^3. A satellite is revolving around near the surface of this planet. Determine the period of revolution of this satellite. (G = 6.67 x 10^-11 Nm^2 kg^-2).

Solution- T = (2π√R^3/GM) = {2π√R^3/(G4/3πR^3p)}
= √3π/Gp
= √(3 x 3.14)/(6.67 x 10^-11 x 8000)
= 4202 sec Ans.

##### Que-28: With what velocity must a body be thrown from earth’s surface so that it may reach a height 4 Re above the earth’s surface ? (Radius of the earth Re = 6400 km, g = 9.8 m/s^2)

Solution-   For body to reach certain height, kinetic energy equals to work done by the body.
Kinetic energy = Work done
1/2 mv^2 = GMm [(1/R – 1/(R+h)]
v^2 = 2GM × h / [R(R+h)]
Substitute h = 4R,
v^2 = 2GM × 4R / [R(R+4R)]
v^2 = 8GM/5R
As GM/R = gR,
v^2 = 8gR/5
v^2 = 8 × 9.8 × 6.4×10^6 / 5
v^2 = 1.003×10^8
v = 10015 m/s
v = 10 km/s Ans.

##### Que-29: A body is thrown vertically upwards with a velocity 10 km/s^2 from the earth’s surface. Up to which height will it go ? Radius of the earth is 6400 km and g = 10m/s^2.

Solution-  v = √2gRe/(1+Re/h)
v^2 = 2gRe/(1+Re/h)
1+Re/h = 2gRe/v^2
Re/h = 2gRe/v^2 – 1 = (2gRe – v^2)/v^2
h = v^2 Re/2gRe – v^2
= {(10 x 10^3)^2 x 6.4 x 10^6}/{2 x 9.8 x 6.4 x 10^6 – (10 x 10^3)^2}
= 2.228 x 10^7 m
= 2.28 x 10^4 km Ans.

##### Que-30: If the radius of the earth be 6.38 x 10^6 m and the acceleration due to gravity at earth be 9.8 m/s^2, then calculate the escape velocity of a body from the earth’s surface.

Solution- Escape velocity
v = √2Rg
= √2×6.38×10^6×9.8
= 11.18×10^3 m/s
= 11.2 km/s Ans.

##### Que-31: Mass of the moon is 7.34 x 10^22 kg and mean radius 1.74 x 10^6 m. If G = 6.67 x 10^-11 Nm^2 kg^-2, then calculate the escape velocity on the moon’s surface.

Solution-   The escape velocity on the moon’s surface can be calculated using the formula:
v = √(2 g r)
Where g is the acceleration due to gravity and r is the radius of the moon.
Substituting the given values:
v = √(2 * 6.67×10⁻¹¹ * 7.34×10²² / 1.74×10⁶)
Simplifying the equation gives v = 2.37 km/h Ans.

##### Que-32: The radius of a planet is four times that of the moon. The value of acceleration due to gravity on the moon is g/5, where g is acceleration due to gravity on the planet. What will be the escape velocity from the planet surface, if its value from the moon surface is 2.5 km/s?

Solution- v = √2gR
Vm/Vp = √2gm Rm/2 gp Rp
2.5/Vp = √gp x Rm / 5gp x 4Rm
2.5/Vp = √1/5 x 1/4 = 1/2 √5
Vp = 2.5 x 2 x √5
= 5√5 = 5 x 2.236 = 11.18 km/s Ans.

##### Que-33: The escape velocity of a projectile on earth’s surface is 11.2 km/s. A body is projected up with twice this speed. What will be the speed of the body at infinity? Ignore the presence of sun and other planets etc.

Solution-  Speed at infinity = ve√n^2 – 1
where n is time of escape velocity by which the object in thrown
v = 11.2 x √2^2 – 1
= 11.2 x √3 = 19.4 km/s Ans.

##### Que-34: The orbital speed of an artificial satellite orbiting very close to the earth is 8 km/s. What should be the increased speed of the satellite so that it may escape leaving its orbit?

Solution- Ve = √2 Vo
Ve = √2 x 8 = 11.3 km/s Ans.

—:  end of Gravitation : Satellites and Escape Velocity Numerical ISC Class 11 Physics Nootan Solutions :—

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