Question

2. Two satellites of masses m and 2m are revolving around a
planet of mass M with different speeds in orbits of radiir
and 2r respectively. The ratio of minimum and maximum forces
on the planet due to satellites is

(a)1/2
(b)1/4
(C)1/3
(d) None of these​

Answer 1

from the figure, centripetal force,

f=

r

mv

2

v

2

=

m

fr

The orbital speed of 1st satellite is

v

1

=

m

fr

. . . .(1)

The orbital speed of 2nd satellite,

v

2

=

2m

f2r

v

2

=

m

fr

. . . . .(2)

Ratio v

1

:v

2

=1:1

The correct option is A.

solution

Answer 2

Given two satellites orbiting the same planet, find the ratio of minimum and maximum forces.

Explanation:

• Let both the satellites be denoted by 's1' and 's2' and forces on the planet due to them be denoted by 'F1' and 'F2' respectively.
• The forces on the planet due to satellites will be minimum when both the satellites are exactly on the opposite side of the planet along a straight line. That is, $F_{min}=F_1-F_2$
• The forces on the planet due to satellites will be maximum when both the satellites are exactly on the same side of the planet along a straight line. That is, $F_{max}=F_1+F_2$    
• Since the mass and the radius of the orbit of 's1' is 'm' and 'r' we get, $F_1=\frac{GMm}{r^2}$  
• Since the mass and the radius of the orbit of 's2' is '2m' and '2r' we get, $F_2=\frac{GMm}{2r^2}$  
• Hence the ratio of minimum and maximum forces is,
•    $R=\frac{F_1-F_2}{F_1+F_2} \\->R=\frac{\frac{GMm}{r^2} (1-\frac{1}{2} )}{\frac{GMm}{r^2}(1+\frac{1}{2} )} \\->R=\frac{1}{3}$      ----------->ANSWER (C)