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\(320 \mathrm{~km}\) above the surface of earth, the value of acceleration due to gravity is nearly \(90 \%\) of its value on the surface of the earth. Its value will be \(95 \%\) of the value on the earth's surface (a) nearly \(160 \mathrm{~km}\) below the earth's surface (b) nearly \(80 \mathrm{~km}\) below the earth's surface (c) nearly \(640 \mathrm{~km}\) below the earth's surface (d) nearly \(320 \mathrm{~km}\) below the earth's surface

Short Answer

Expert verified
Option (d): nearly 320 km below the Earth's surface.

Step by step solution

01

Understanding gravity above the surface

The acceleration due to gravity at a height above the Earth's surface, like 320 km, is given by the formula \( g_h = g_0 \left(\frac{R}{R+h}\right)^2 \) where \( g_0 \) is the gravitational acceleration on the surface, \( R \) is the Earth's radius, and \( h \) is the height above the surface. For 320 km, \( g_h = 0.9g_0 \). This is our reference point.
02

Understanding gravity below the surface

The acceleration due to gravity below the Earth's surface can be approximated by \( g_d = g_0 \left(1- \frac{d}{R} \right) \) where \( d \) is the depth below the surface. At this point, we know \( g_d = 0.95 g_0 \).
03

Set up the equation for below the surface

Using the formula \( g_d = g_0 \left(1- \frac{d}{R} \right) = 0.95 g_0 \), we can set up the equation \( 1- \frac{d}{R} = 0.95 \).
04

Solve for the depth (d)

From the equation \( 1 - \frac{d}{R} = 0.95 \), we rearrange to find \( \frac{d}{R} = 0.05 \). Therefore, \( d = 0.05R \). Assuming \( R \approx 6400 \text{ km} \), the depth \( d \approx 0.05 \times 6400 = 320 \text{ km} \).
05

Identify the correct option

The calculated depth of 320 km below the Earth's surface matches the condition where gravity is 95% of its surface value. Hence, option (d) nearly matches our calculation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

gravity variations
Gravity doesn't remain constant across different locations on Earth and changes based on several factors. These variations can happen due to changes in elevation, latitude, and local geological structures. When you are above or below the Earth's surface, the gravitational pull could be weaker or stronger, respectively.
- **Above the surface**: Higher altitudes, such as those experienced in high mountains or airplanes, mean you are further from Earth's center. Here, gravity weakens and you weigh a tiny bit less.
- **Below the surface**: When you dig deeper into Earth, you get closer to the massive central core. As a result, gravity can become stronger because there is more mass above you exerting a pull.

Understanding these gravity variations helps explain why certain locations on Earth have slightly different gravitational forces. This is crucial for applications in geophysics and engineering.
Earth's radius
The radius of Earth is an essential parameter for calculating gravitational acceleration. Earth's average radius is approximately 6400 kilometers. This value is fundamental for applying the gravitational formula to predict how gravity changes with altitude and depth.
- At **sea level**, the radius is typically measured from the center of Earth to that point on the surface.
- When you travel **above** the surface, the radius becomes the distance from Earth's center to the point of interest above the surface.
- Similarly, **under the surface**, the radius still extends from the center to where you are below the surface level.

Using Earth's radius in our calculations helps determine how the gravitational force diminishes with height or increases with depth. This knowledge is utilized by scientists and engineers in fields such as satellite launching and underground exploration.
gravitational formulas
Gravitational formulas play a vital role in evaluating how gravity differs with position relative to Earth. The two key formulas are for calculating gravity above and below the surface.

### Above the SurfaceThe formula to calculate gravity above the surface is \[ g_h = g_0 \left(\frac{R}{R+h}\right)^2 \] where:
  • \(g_h\) is the gravity at height \(h\)
  • \(g_0\) is the gravity at Earth's surface
  • \(R\) is Earth's radius, and
  • \(h\) is the height above the surface
This shows that as height increases, gravity decreases.

### Below the SurfaceWhen you're below the surface, gravity is computed using the formula:\[ g_d = g_0 \left(1 - \frac{d}{R} \right) \] with:
  • \(g_d\) as gravity at depth \(d\)
  • \(g_0\) as surface gravity
  • \(d\) as depth beneath the surface
Here, as the depth increases, the gravitational force becomes a function of how close you are to the center through the depth fraction.By understanding these formulas, students can better visualize how gravitational force changes around Earth, enhancing comprehension in physics.

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Most popular questions from this chapter

A comet of mass \(m\) moves in a highly elliptical orbit around the sun of mass \(M\). The maximum and minimum distances of the comet from the centre of the sun are \(r_{1}\) and \(r_{2}\) respectively. The magnitude of angular momentum of the comet with respect to the centre of sun is (a) \(\left[\frac{G M r_{1}}{\left(r_{1}+r_{2}\right)}\right]^{1 / 2}\) (b) \(\left[\frac{G M m r_{1}}{\left(r_{1}+r_{2}\right)}\right]^{1 / 2}\) (c) \(\left(\frac{2 G m^{2} r_{12}}{r_{1}+r_{2}}\right)^{1 / 2}\) (d) \(\left(\frac{2 G M m^{2} r_{12}}{\left(r_{1}+r_{2}\right)}\right)^{1 / 2}\)

The work that must be done in lifting a body of weight \(P\) from the surface of the earth to a height \(h\) is (a) \(\frac{P R h}{R-h}\) (b) \(\frac{R+h}{P R h}\) (c) \(\frac{P R h}{R+h}\) (d) \(\frac{R-h}{P R h}\)

Two equal masses \(m\) and \(m\) are hung from a balance whose scale pan differs in vertical height by \(h / 2\). 'The error in weighing in terms of density of the earth \(\rho\) is (a) \(\frac{1}{3} \pi G \rho m h\) (b) \(\pi G \pi m h\) (c) \(\frac{4}{3} \pi G \rho m h\) (d) \(\frac{8}{3} G \rho m h\)

What should be the angular speed of earth in \(\mathrm{rads}^{-1}\) so that a body of \(5 \mathrm{~kg}\), weighs zero at the equator? (Take \(g=10 \mathrm{~ms}^{-2}\) and radius of earth \(=6400 \mathrm{~km}\) ). (a) \(1 / 1600\) (b) \(1 / 800\) (c) \(1 / 400\) (d) \(1 / 80\)

The gravitational field due to a mass distribution is \(1=\frac{C}{x^{2}}\) in \(x\) direction. Here \(C\) is constant. Taking the gravitational potential to be zero at infinity, potential at \(x\) is (a) \(\frac{2 C}{x}\) (b) \(\frac{C}{x}\) (c) \(\frac{2 C}{x^{2}}\) (d) \(\frac{C}{2 x^{2}}\)

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