0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
Question

If g on the surface of the Earth is 9.8 ms2, then it's value at a depth of 3200 km (Radius of the earth =6400 km) is
  1. 9.8 ms2
  2. 2.45 ms2
  3. zero
  4. 4.9 ms2

A
4.9 ms2
B
9.8 ms2
C
zero
D
2.45 ms2
Solution
Verified by Toppr

The value of gravity changes as we move away from or towards the centre of the Earth.
This is given by: gR=GMR2, where M is the mass of a planet of radius R
So, M=43πR3ρ ; substituting in above equation
gR=G43πρ×R
Since we want the value of g at depth(d) from the Earth's surface, we replace R by (Rd)
gd=G43πρ×(Rd)
gRgd=RRd
gdgR=(1dR)

Was this answer helpful?
10
Similar Questions
Q1
If g on the surface of the Earth is 9.8 ms2, then it's value at a depth of 3200 km (Radius of the earth =6400 km) is
View Solution
Q2
If g on the surface of the Earth is 9.8 ms2, its value at a height of 6400km is: (Radius of the Earth =6400km)

View Solution
Q3
The value of g at a particular point on the earth surface is 9.8 ms2. Suppose the earth suddenly shrinks uniformly to half its present size without losing any mass. The value of ‘g’ at the same point (assuming that the distance of the point from the centre of earth does not shrink) will now be:
View Solution
Q4
Calculate the acceleration due to gravity at a point
a] 64 km above and
b] 32 km below the surface of earth.
Given Radius of Earth =6400 km. Acceleration due to gravity at the surface of earth =9.8 ms2.
View Solution
Q5
Assume that a satellite is revolving around Earth in a circular orbit almost close to the surface of Earth. The time period of revolution of satellite is (Radius of earth is 6400 km, g= 9.8 ms2)
View Solution