Gravitation

A satellite is launched into a circular orbit of radius R round the earth. A second satellite is launched into an orbit of radius (1.01)R. the period of the second satellite is larger than that of the first one by approximately

Q: A satellite is launched into a circular orbit of radius R round the earth. A second satellite is launched into an orbit of radius (1.01)R . The period of the second satellite is larger than that of the first one by approximately

(a) 0.5 %

(b) 1.0 %

(c) 1.5 %

(d) 3.0 %

Click to See Answer :
Ans: (c)
Sol: $T^2 \propto R^3 $

$(\frac{T’}{T})^2 = (\frac{1.01 R}{R})^3 $

$(\frac{T’}{T}) = (\frac{1.01 R}{R})^{3/2} $

$\frac{T’}{T} = (1 + \frac{1}{100})^{3/2} $

$\frac{T’}{T} = (1 + \frac{3}{2}\frac{1}{100}) $ (Using Binomial Theorem )

$\frac{T’}{T} -1 = (\frac{3}{2}\frac{1}{100}) $

$\frac{T’ – T}{T} = (\frac{3}{2} \times \frac{1}{100}) $

$(\frac{T’-T}{T} )\times 100 = (\frac{3}{2}\times \frac{1}{100}) \times 100 $

= 1.5 %

 

If r denotes the distance between the sun and the earth, then the angular momentum of the earth around the sun is proportional to

Q: If r denotes the distance between the sun and the earth, then the angular momentum of the earth around the sun is proportional to

(a) r3/2

(b) r

(c) √r

(d) r2

Click to See Answer :
Ans: (c)
Sol: Angular Momentum , L = m v r

$L = m \sqrt{\frac{G M}{r}} r $

$L \propto \sqrt{r} $

 

How long will a satellite, placed in a circular orbit of radius that is (1/4)^th the radius of a geostationary satellite, take to complete one revolution around the earth

Q: How long will a satellite, placed in a circular orbit of radius that is (1/4)th the radius of a geostationary satellite, take to complete one revolution around the earth

(a) 12 hours

(b) 6 hours

(c) 3 hours

(d) 4 days

Click to See Answer :
Ans: (c)
Sol: As , $T^2 \propto r^3 $

$ (\frac{T’}{T})^2 = (\frac{r’}{r})^3 = (\frac{r/4}{r})^3 $

$ (\frac{T’}{T})^2 = \frac{1}{64} = \frac{}{}$

$ (\frac{T’}{T}) = \frac{1}{8} $

T’ = 3 hours

 

In planetary motion the areal velocity of position vector of a planet depends on angular velocity (ω) and the distance of the planet from sun (r). If so the correct relation for areal velocity is

Q: In planetary motion the areal velocity of position vector of a planet depends on angular velocity (ω) and the distance of the planet from sun (r). If so the correct relation for areal velocity is

(a) $\frac{dA}{dt} \propto \omega r$

(b) $\frac{dA}{dt} \propto \omega^2 r$

(c) $\frac{dA}{dt} \propto \omega r^2 $

(d) $\frac{dA}{dt} \propto \sqrt{\omega r} $

Click to See Answer :
Ans: (c)
Sol: $\frac{dA}{dt} = \frac{d}{dt} (\frac{1}{2}) r (r \theta ) $

$\frac{dA}{dt} = (\frac{1}{2}) r^2 (\frac{d \theta }{dt} ) $

$\frac{dA}{dt} = (\frac{1}{2}) r^2 \omega $

 

A satellite close to earth is in orbit above the equator with a period of rotation of 1.5 hrs. If it is above a point P ….

Q: A satellite close to earth is in orbit above the equator with a period of rotation of 1.5 hrs. If it is above a point P on the equator at some time, it will be above P again after time

(A) 1.5 hrs.

(B) 1.6 hrs. if it is rotating from west to east.

(C) 24/17 hrs. if it is rotating from west to east.

(D) 24/17 hrs. if it is rotating from east to west.

Click to See Answer :

Solution : Let ωo = the angular velocity of earth above its axis = 2π/24  rad/hr

Let ω = angular velocity of satellite.

ω = 2π/1.5

For a satellite rotating from west to east. (same as earth), the relative angular velocity ω1 = ω – ωo

Time period of rotation relative to earth = 2π/ωω1 = 1.6 h

Now for a satellite rotating from east to west (opposite to earth) the relative angular velocity ω2 = ω + ωo

Time period of rotation relative to earth =2π/ω2 = 24/17 hrs.