The footfall player has a mass equal to 100kg standing on earth surface at distance of 6.38×10^6m.calculate force of gravitational attraction between the earth and football player?

Answer 1

#approx 1000N#

Applying the universal gravitational law of Newton:

#F=G(Mm)/(r^2)#

Given two masses' respective masses and their proximity to one another, we can find the attraction force between them.

The mass of the football player is #100kg# (let's call it #m#), and the mass of the Earth is #5.97 times 10^24# kg.(let's call it #M#).
And as the distance should be is measured from the center of the object, the distance the Earth and the player are from each other must be Earth's radius- which is the distance given in the question- #6.38 times 10^6# meters.
#G# is the gravitational constant, which has a value of #6.67408 × 10^-11 m^3 kg^-1 s^-2#

Let's now enter everything into the formula:

#F=(6.67408 times 10^-11) times ((100) times (5.97 times 10^24))/(6.38 times 10^6)^2#
#F=978.8N approx 1000N# as the least amount of significant figures given is 1 significant figure.
This closely resembles the value of the gravitational field strength or Earth, #g#.

The gravitational field strength, or force per unit mass, can be found using the following equation:

#g=(F)/m#
We can test our answer. In reality, #g=9.81 ms^-2#

In light of our value:

#g=978.8/100#
#g=9.788 approx 9.81#

Thus, it appears to be in order.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

The force of gravitational attraction between the Earth and the football player can be calculated using Newton's law of universal gravitation formula:

F = (G * m1 * m2) / r^2

Where: F = force of gravitational attraction G = gravitational constant (6.674 × 10^-11 N*m^2/kg^2) m1 = mass of the Earth (5.972 × 10^24 kg) m2 = mass of the football player (100 kg) r = distance between the centers of mass of the Earth and the football player (6.38 × 10^6 m)

Plugging in the values:

F = (6.674 × 10^-11 * 5.972 × 10^24 * 100) / (6.38 × 10^6)^2

F ≈ (6.674 × 10^-11 * 5.972 × 10^26) / (4.0684 × 10^13)

F ≈ 9.81 N

So, the force of gravitational attraction between the Earth and the football player is approximately 9.81 Newtons.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7