Kepler's third law states that square of the period of revolution (T) of a planet around the sun, is proportional to the third power of average distance r between the sun and planet i.e, T2 =Kr3, here K is constant.
If the masses of the sun and planet are M and m respectively, them as per Newton's law of gravitation force of attraction between them is
The relation between G and K is described as
GK =4π2
GMK =4π2
K=G
K=G
B.
GMK =4π2
The gravitational force of attraction between the planet and sun provide the centripetal force
Variation of acceleration due to gravity (g) with distance x from the centre of the Earth is best represented by (R → Radius of the Earth)
D.
The value of g is maximum at the surface of the Earth and g = 0 at centre of the Earth. If one goes away from the Earth surface, again the value of g decreases. Thus, graph (d) is the correct variation of g.
At what height from the surface of earth the gravitation potential and the value of g are -5.4 x 107 J kg-2 and 6.0 m/s2 respectively? (Take the radius of the earth as 6400 km)
1600 km
1400 km
2000 km
2600 km
D.
2600 km
Gravitational potential at some height h from the surface of the earth is given by,
V = ... (i)
Acceleration due to gravity at some height h from the earth surface can be given as,
... (ii)
From equation (i) and (ii), we get
... (iii)
V = -54 x 107 J kg-2
g' = 6.0 m/s2
Radius of Earth, R = 6400 km
The value of acceleration due to gravity at a depth of 1600 km is equal to
4.9 ms-2
9.8 ms-2
7.35 ms-2
19.6 ms-2
C.
7.35 ms-2
Given,
Depth (d) = 1600 km
We know that,